## Zeros of Dedekind zeta functions in the critical strip

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Venue: | Math.Comp.66 (1997), 1295–1321. MR 98d:11140 Laboratoire d’Algorithmique Arithmétique, Université BordeauxI,351coursdela Libération, F-33405 Talence Cedex France E-mail address: omar@math.u-bordeaux.fr |

Citations: | 6 - 0 self |

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@INPROCEEDINGS{Tollis_zerosof,

author = {Emmanuel Tollis},

title = {Zeros of Dedekind zeta functions in the critical strip},

booktitle = {Math.Comp.66 (1997), 1295–1321. MR 98d:11140 Laboratoire d’Algorithmique Arithmétique, Université BordeauxI,351coursdela Libération, F-33405 Talence Cedex France E-mail address: omar@math.u-bordeaux.fr},

year = {}

}

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### Abstract

Abstract. In this paper, we describe a computation which established the GRH to height 92 (resp. 40) for cubic number fields (resp. quartic number fields) with small discriminant. We use a method due to E. Friedman for computing values of Dedekind zeta functions, we take care of accumulated roundoff error to obtain results which are mathematically rigorous, and we generalize Turing’s criterion to prove that there is no zero off the critical line. We finally give results concerning the GRH for cubic and quartic fields, tables of low zeros for number fields of degree 5 and 6, and statistics about the smallest zero of a number field. 0. Introduction and notations The Riemann zeta function and its generalization to number fields, the Dedekind zeta function, have been for well over a hundred years one of the central tools in number theory. It is recognized that the deepest single open problem in mathematics is the settling of the Riemann Hypothesis, and number theorists know that its

### Citations

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Handbook of Mathematical Functions with Formulas
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- 1992
(Show Context)
Citation Context ...− r2 ln 1 − 2j j=1 X � j = � X k≥1 k � (−1) k k � � r1 ζ(k) + r2 + 2k r1 2k Hq,k � + r2H2q,k � − r1 ln (−1) q q! X � − r2 ln[(2q)!X] 2sZEROS OF DEDEKIND ZETA FUNCTIONS IN THE CRITICAL STRIP 1299 (see =-=[1]-=- for the expansion formula of ln Γ(1+z)). To prove (12), we let X = z+2q+1 and we have � � X − 2q − 1 Γ(X +1) Γ = 2 √ π � X 2 −q−1� ... � X 2 −1� Γ � X 2 +1�2X−2q−2 (X−2q−2) ...(X−1) . We compute the ... |

261 |
Algebraic Number Theory
- Lang
- 1984
(Show Context)
Citation Context ...mula (16), p. 431), to prove the following inequality: � +∞+it � 3 2 ln |ζK(s)| ds ≥ +it � � � ln � ζK(s) � � �ζK(s +1) �ds − 1.2183N. (27) 1 2 +it 1 2 +it We apply the following result about ΛK (see =-=[11]-=-). Lemma 4.3. The function s → s(s−1)ΛK(s) is holomorphic on the complex plane and its zeros are the non-trivial zeros ρ of the Dedekind zeta function ζK. So ΛK can be expanded in a Weierstrass-Hadama... |

221 |
Riemann’s zeta function
- Edwards
- 1974
(Show Context)
Citation Context ... ∂R ′ K (s)ds = LK 1 π Im �� L C ′ � K (s)ds LK = 1 π Im �� d � � s � � � r1 ln Γ + r2 ln Γ(s)+sln CK ds C ds 2 + 1 π Im �� � d [ln s(s − 1)] ds + ds 1 π Im �� ζ ′ � K (s)ds C C ζK and we obtain (see =-=[6]-=-, p. 128 for a rigorous justification) the relation NK(T )= ΦK(T) (32) +1+SK(T). π We can easily compute an approximation of ΦK, so the main problem is to estimate the function SK(t). Remark. We can d... |

60 |
Vorlesungen über Differenzenrechnung
- Nörlund
- 1924
(Show Context)
Citation Context ...end only on the numbers1300 EMMANUEL TOLLIS field and on the absolute precision required). If s is not an integer ΛK(s) = 2r1hKRK ωK(s−1)s + Cs � s �r1 KΓ Γ(s) 2 r2 N0 � ann −s (14) + SK(s)+SK(1 − s) =-=(15)-=- (16) r1+r2−1 � SK(s) =− N0 � ci,k = n=1 k=0 r1+r2−k � j=1 + C 1−s K Γ i0� i=0 � 1 − s 2 ci,k , (s + i) k+1 � n anAi,j+k CK � r1 � i (ln CK n=1 Γ(1 − s) r2 N0 � ann s−1 + ɛ, where n )j−1 . (j − 1)! Pr... |

41 |
Bounds for discriminants and related estimates for class numbers, regulators and zeros of zeta functions: a survey of recent results
- Odlyzko
- 1990
(Show Context)
Citation Context ...nly on the numbers1300 EMMANUEL TOLLIS field and on the absolute precision required). If s is not an integer ΛK(s) = 2r1hKRK ωK(s−1)s + Cs � s �r1 KΓ Γ(s) 2 r2 N0 � ann −s (14) + SK(s)+SK(1 − s) (15) =-=(16)-=- r1+r2−1 � SK(s) =− N0 � ci,k = n=1 k=0 r1+r2−k � j=1 + C 1−s K Γ i0� i=0 � 1 − s 2 ci,k , (s + i) k+1 � n anAi,j+k CK � r1 � i (ln CK n=1 Γ(1 − s) r2 N0 � ann s−1 + ɛ, where n )j−1 . (j − 1)! Proof. ... |

25 | Numerical Computations Concerning the ERH
- Rumely
- 1993
(Show Context)
Citation Context ...tant. Much energy has been devoted to the numerical investigation of the zeta function (see [4] for example). There has been some investigations of its closest cousins, the Dirichlet L-functions (see =-=[20]-=-). However, the case of a general number field has remained totally unexplored territory. We give here the first numerical evidence in favour of the Generalized Riemann Hypothesis for a number field w... |

18 |
Asymptotic expansions and analytic continuations for a class of Barnesintegrals
- Braaksma
- 1963
(Show Context)
Citation Context ...fy the multiple integral appearing in Hecke’s formula by turning it into a single integral. The main result that we will need is the following (see Proposition 2.3 and Theorem 1 in [7]). Theorem 1.1. =-=(3)-=- ΛK(s) = 2r1 hKRK ωK(s−1)s + � with the notations of Section 0 and (4) f(x, s) = 1 2iπ � δ+i∞ δ−i∞ x z Γ n≥1 an CK = � � CK f n ,s � � CK +f n ,1−s �� � |DK| π N 2 2r2 , � z �r1 dz r2 Γ(z) , δ > max(R... |

15 |
Analytic formulas for the regulator of a number field
- Friedman
- 1989
(Show Context)
Citation Context ... � −i0− 1 2 � � �� Γ �z� ���� � � z ��r1 �� dz r2 �Γ |Γ(z)| 2 z − s � −i0 − 1 � 2 + it 2 ��r1 � ��� ��� Γ −i0 − 1 + it 2 ��r2 ��� dt.sZEROS OF DEDEKIND ZETA FUNCTIONS IN THE CRITICAL STRIP 1305 As in =-=[8]-=-, we have � � � �Γ � −i0 − 1 �� � ��� �Γ + it ≤ 2 � 1 2 + it��� � � � −i0 − 1 2 + it��−i0 + 1 2 +it� ... � −1 2 +it��� ≤ 2 � � � i0! �Γ � �� 1 ��� + it 2 and � � � �Γ � −i0 − 1 �� 2 + it ��� ≤ 2 4 � �... |

12 |
On the Phragmén-Lindelöf theorem and some applications, Math. Zeitschrift 72
- Rademacher
- 1959
(Show Context)
Citation Context ... have (23) IK(t) = � +∞+it 1 2 +it ln |ζK(s)| ds � � � � N t ≤ (0.8252 + 2.329N)+0.1407 ln |DK| . 2π Proof. The proof of this theorem is quite easy. We apply the following result from Rademacher (see =-=[18]-=- p. 200).sZEROS OF DEDEKIND ZETA FUNCTIONS IN THE CRITICAL STRIP 1307 Lemma 4.2. If 0 <η< 1 2 and s = σ + it, we have with the usual notations � � � �1+η−σ � �N 2 � |ζK(s)| ≤3� 1+s� � |1+s| (24) �1−s�... |

11 |
On computing Artin L-functions in the critical strip
- Lagartas, Odlyzko
- 1979
(Show Context)
Citation Context ...he relation NK(T )= ΦK(T) (32) +1+SK(T). π We can easily compute an approximation of ΦK, so the main problem is to estimate the function SK(t). Remark. We can directly use the argument principle (see =-=[10]-=-, p 1093). Indeed, the argument principle asserts the number of zeros of LK in R is the change in argument of this function around ∂R divised by 2π. Using the symmetries of ΛK, we have only to conside... |

8 |
On the distribution of the zeros of the Riemann zeta-function
- Lehman
- 1970
(Show Context)
Citation Context ...rsively computed. For a given value of s, wecannow deduce the coefficients Ai,j(s) from the preceding results. Formulas (7) and (10) give immediately the following relations between Ai,j and Ai,j(s). =-=(13)-=- if i �= −s Ai,j(s) = Ai,j+1(s)−Ai,j � s+i Ai,r1+r2+1(s) =0 ifiis even, Ai,r2+1(s) =0 ifiis odd, if i = −s A−s,j(s) =A−s,j−1. with 2.3. Another form of the method. Our first goal in this paper is to c... |

7 |
Sur les zéros de la fonction ζ(s) de
- Backlund
- 1914
(Show Context)
Citation Context ...al to n. This function is extended by analytic continuation to a meromorphic function on the whole complex plane with a unique simple pole at s = 1. Moreover, ζK verifies a functional equation: where =-=(2)-=- ΛK(s) =Γ ΛK(s) =ΛK(1 − s) � s �r1 Γ(s) 2 r2 �� |DK| π N 2 2 r2 � s ζK(s) and ΛK is meromorphic on the complex plane with two poles at s =1andats=0. 1. The method for computing values In [7], Eduardo ... |

6 |
On the Zeros of the Riemann Zeta Function
- Brent, Winter
- 1982
(Show Context)
Citation Context ...to a single integral. The main result that we will need is the following (see Proposition 2.3 and Theorem 1 in [7]). Theorem 1.1. (3) ΛK(s) = 2r1 hKRK ωK(s−1)s + � with the notations of Section 0 and =-=(4)-=- f(x, s) = 1 2iπ � δ+i∞ δ−i∞ x z Γ n≥1 an CK = � � CK f n ,s � � CK +f n ,1−s �� � |DK| π N 2 2r2 , � z �r1 dz r2 Γ(z) , δ > max(Re s, 0). 2 z − s Apart from the problem of computing the regulator and... |

6 |
Analytic computations in number theory
- Odlyzko
- 1994
(Show Context)
Citation Context ...∞ x 2iπ −z � 1 h2(z,s)dz = H0 x ,s � . −δ−i∞ We need an asymptotic bound for H0( 1 x ,s)whenx→0+ . We have obtained the following result. Proposition 3.1. If 0 <x≤1,|Im s| ≤T,and|Re s| ≤L, we have: � =-=(17)-=- with |f(x, s)| ≤Ax r1 +r2−3 � N exp −2 r1 N −1 − Nx 2 N A = π r1 +r2 2 2 N 1 2 r1+r2− N 2 (r2 + LN +1) r2 +LN+1 2 (T + L +1). Proof. We first apply the Euler-Maclaurin formula to obtain an estimate f... |

5 |
Hecke’s integral formula. Séminaire de Théorie des Nombres Univ
- Friedman
- 1987
(Show Context)
Citation Context ...on: where (2) ΛK(s) =Γ ΛK(s) =ΛK(1 − s) � s �r1 Γ(s) 2 r2 �� |DK| π N 2 2 r2 � s ζK(s) and ΛK is meromorphic on the complex plane with two poles at s =1andats=0. 1. The method for computing values In =-=[7]-=-, Eduardo Friedman explains how we can simplify the multiple integral appearing in Hecke’s formula by turning it into a single integral. The main result that we will need is the following (see Proposi... |

3 |
On the distribution of the zeros of the Riemann zeta function
- Lehman
- 1970
(Show Context)
Citation Context ...this section. The first one is [21] (in 1953) where Turing explains his method. Unfortunately, there are many mistakes in this paper but Lehman gives a corrected version of Turing’s original proof in =-=[12]-=-. Finally, we refer to a third paper: in [20], Rumely generalizes the criterion to Dirichlet L-series. We introduce the function SK(t) = 1 1 arg ζK( + it) π 2 where the value of the argument is obtain... |

2 |
die Zetafunktion beliebiger algebraischer
- Hecke, Uber
- 1917
(Show Context)
Citation Context ...als pi (i =1,...,m) such that N pi ≤ N0. Let fi= fpi be the residual degree of pi and pi the prime number below pi. We compute the an recursively, introducing the coefficients an,h (for h =0,...,m)by =-=(9)-=- � a1,0=1 and an,0 =0 forn≥2, if h ≥ 1. �∞ n=1 an,hn−s = �h � 1 i=1 k≥0 p kfis i With a little computation, we have an,h = v� k=0 a n P k ,h−1 where P = p fh h ,andvis the largest integer such that P ... |

1 |
Etude de la fonction nombre de façons de représenter un entier comme produit de k facteurs, Doctorat de l’Université de Limoges
- Duras
- 1993
(Show Context)
Citation Context ... to find the “best” value of N0 such that � |r(n, s)| = n>N0 � � � an � � n>N0 f � CK n ,s � � CK +f n ,1−s �� ��� ≤ɛ. We define the coefficients dN (n) byζ(s) N = � n≥1dN(n)n−sand using Theorem 2 in =-=[5]-=-, we obtain the inequality (18) From this and (17) we can derive and where K0 = C Nb(δ+1)N 2 2 an ≤ dN (n) ≤ n a where a = |r(n, s)| ≤2AC α−3 3−α µ K na+ µ exp � (δ+1)N−2 e � −n 2 N C 2 N K (ln N)3 ln... |

1 |
Theorie der Gamma-Funktion
- Nielsen, Der
- 1965
(Show Context)
Citation Context ...omplex number with |Im s| ≤T and |Re s| ≤L.Weletµ=N/2, β =2r1/2and α =(r1+r2+3)/2. We introduce the function g(z,s)= (2π)1−r1−r2 h2(z,s)(βµ µ ) −z . Γ(1 − α − µz) We start with the following formula (=-=[14]-=-, p 208, formula (17)): ln Γ(z + a) =(z+a)lnz−z+ 1 ln 2π + φ(z) 2 with φ(z) = Q(z) and |Q(z)| ≤2ifRez>0. 12zs1302 EMMANUEL TOLLIS We use this result to compute ln g(z,s)=−zln(βµ µ ) − ln(−s − z)+(1−r1... |

1 |
Some calculations of the Riemann zeta function, Proc
- Turing
- 1953
(Show Context)
Citation Context ...esult depends on the degree, the signature and the discriminant of the number field. Theorem 4.1. If t1 and t2 are real numbers such that 40 <t1 <t2, we have � � 1 SK (t2) − S 1 K (t1) � � � � � N t2 =-=(21)-=- � ≤ (0.2627 + 1.8392N)+0.122 ln DK . 2π Proof. We assume that t1 > 40 and t2 > 40 since in the case of number fields of degree 4, we compute zeros to height 40. We note that there is a constant term ... |