André Leclair Cornell University

THe Riemann Hypothesis : a New perspective

AndrÉ LeClair Cornell university

Paris, May 11, 2015 (LPTHE) Outline • Riemann Zeta in Quantum Statistical Physics. • Riemann Hypothesis • Zeta and the distribution of Prime Numbers. • Zeta and Random Matrix Theory. • Generalized RH for Dirichlet and modular L- functions • 3 Conjectures and 2 concrete strategies toward a proof.

Introductory material reviewed in my Riemann Center lectures: arXiv:1407.4358 Preface

• Main results involve an interplay between the Euler Product Formula and the functional equation. • Conjectures are “derived” but we do not prove rigorous mathematical theorems. • The approach is universal, i.e. applies to at least two inﬁnite classes of “zeta” functions. • It’s a “constructive” approach, i.e. leads to new formulas, etc. Riemann Zeta Function was present 3 Numericalat the Values birth of Quantum Mechanics: 11. The values of both universal constants h and k may be calculated rather precisely with the aid of § available measurements. F. Kurlbaum14, designating the total energy radiating into air from 1 sq cm of a blackOn body the at Law temperature of DistributiontC in 1 sec by St of, found Energy that: in the Normal Spectrum

watt 5 erg S100 S0 = 0.0731Max Planck=7.31 10 · cm2 · · cm2 sec · From this one can obtain theAnnalen energy density der Physik, of the total vol. radiation 4, p. 553 energy↵ (1901) in air at the absolute temperature 1: 5 4 7.31 10 15 erg · · =7.061 10 This PDF ﬁle was typeset3 with1010 LAT(373X based4 273 on4) the HTML· ﬁle at · cm3 deg4 · · E · Onhttp://dbhs.wvusd.k12.ca.us/webdocs/Chem-History/Planck-1901/Planck-1901.html. the other hand, according to equation (12) the energy density of the total radiant energy for = 1 is: Please report typos etc. to Koji Ando at Kyoto University ([email protected]). 3 1 8⌅h 1 ⇤ d⇤ u⇤ = ud⇤ = Bose-Einstein distribution c3 eh⇤/k 1 ⇤0 ⇤0 8⌅h 1 3 h⇤/k 2h⇤/k 3h⇤/k = ⇤ (e + e + e + )d⇤ c3 ··· ⇤0 and by termwise integration:

8⌅h k 4 1 1 1 u⇤ = 6 1+ + + + c3 · h 24 34 44 ··· ⇥ ⇥ 48⌅k4 = 1.0823 c3h3 · 15 10 If we set this equal to 7.061 10 , then, since c =3 10 cm/sec, we obtain: · TEX for keynote· k4 =1.1682 1015 (14) A very bad typo of the English translation.h3 ·Should read: 4 12. O. Lummer and1 E. Pringswim1 1 15 determined the product⇥ ⇥ , where ⇥ is the wavelength of § 1+ + + + ..... = (4) = =1m .0823 m maximum energy in air4 at temperature4 4 , to be 2940 micron degree. Thus, in absolute measure: 2 3 4 ·90 ⇥ =0.294 cm deg m · On the other hand, it follows from equation (13), when one sets the derivative of E with respect to equal to zero, thereby ﬁnding ⇥ = ⇥m ch 1 ech/k⇥m =1 5k⇥ · m ⇥ and from this transcendental equation: ch ⇥ = m 4.9651 k · consequently: h 4.9561 0.294 11 = · = 4.866 10 k 3 1010 · · From this and from equation (14) the values for the universal constants become:

27 h = 6.55 10 erg sec (15) · ·

16 erg k = 1.346 10 (16) · · deg These are the same number that I indicated in my earlier communication.

14F. Kurlbaum, Wied. Ann. 65 (1898), p. 759. 15O. Lummer and Pringsheim, Transactions of the German Physical Society 2 (1900), p. 176.

1 1 1 1 (z)=1+ + + + ... TEX for keynote 2z 3z 4z 1 1 1 1 (z)= + + + ... 2z 2z 4z 6z 1 1 1 ⇥4 1+The4 + Riemann4 + 4 + ..... Zeta= (4) Function = =1.0823 1 1 1 2 3 4 90 (1 )(z)=1+ + + ... TEX for keynote 2z 3z 5z

1 1 1 1 1 1 1 1 4 (1 )(1 )(z)=1+ + + ... 1 1 1 ⇥ 3z 2z 5z 7z (z)=1+ +=1++ +TEX.....+ = for+(4)..., keynote = =1.0823(z) > 1 2n4 z 34 442z 3z 90 < n=1 1 X (z)= ,p=prime 1 p z 4 p 11 1 1 11 1 ⇥ It can be analytically(z1+)= continued+ =1++ to the++ whole.....+ =..., complex(4) =(z) >z=plane.=11 .0823 For example, 4 nz 4 24z 3z < 1 by considering “fermions”:n2=1 3 4 90 n th zero on critical line : ⇥n = + iyn X 2 z 1 1 1 t 1 ((n +1)=n!) (z)= 1 1 z dt1 t 1 , (z) > 0 (z)=(z)(1 2 =1+ ) 0 e++1 + ...,< (z) > 1 nz Z 2z 3z < n=1 X z 1 1 1 t (z)= dt , (z) > 0 (z)(1 21 z) et +1 < Z0

Trivial zeros: ( 2) = ( 4) = ( 6) ...=0

1 these are invariant if the partition functions are multiplied by a common factor as induced by a global shift of the energies. Based on his understanding of quantum electrodynamics and his own treatment of the Casimir eect, Schwinger once said [7], “...the vacuum is not only the state of minimum energy, it is the state of zero energy, zero momentum, zero angular momentum, zero charge, zero whatever.” A quantum consequence of this for instance is the fact that photons do not scatter o the vacuum energy. All of thisthese strongly are invariant suggests if the partition that functions it is are impossible multiplied by a to common harness factor as vacuum energy in order to do work,induced which by a global in turn shift of calls the energies. into question Based on his whether understanding it of could quantum electrodynamics and his own treatment of the Casimir eect, Schwinger once said be a source of gravitation. [7], “...the vacuum is not only the state of minimum energy, it is the state of zero energy, zero momentum, zero angular momentum, zero charge, zero whatever.” A The Casimir eect is often correctly cited as proof of the reality of vacuum energy. quantum consequence of this for instance is the fact that photons do not scatter However it needs to be emphasizedo the that vacuum what energy. is All actually of this strongly measured suggests that is it the is impossiblechange to harnessof vacuum energy in order to do work, which in turn calls into question whether it could the vacuum energy as one varies a geometric modulus, i.e. how it depends on this be aTEX source forof gravitation. keynote modulus, and this is unaected by anThe Casimirarbitrary eect is shift often correctly of the cited zero as proof of energy. of the reality The of vacuum classic energy. However it needs to be emphasized that what is actually measured is the change of experiment is to measureThe Casimir the forcethe e betweenff vacuumect energy and two as Zeta plates one varies as a geometric one changes modulus, their i.e. how separation; it depends on this modulus, and this is unaected by an arbitrary shift of the zero of energy. The classic the modulus in question here is theseparation distance = ⌥ ⌅ between the plates and the force experiment is to measure the force between two plates as one changes their separation; depends on how the vacuum energythe modulus varies in question with this here is separation. the distance ⌅ between The the Casimir plates and force the force N N dependsN onN how thed3 vacuumk energy varies with this separation.k4 The Casimir force b f b f ⇤ 2 2 c F (⌅)isminusthederivativeoftheelectrodynamicvacuumenergy⇥vac = ⇧k = 3 k + m (Nb Nf )Evac2(⌅)between 2 F (⌅)isminusthederivativeoftheelectrodynamicvacuumenergy2 (2) ⇥ 16 Evac(⌅)between k Z X the two plates, F (⌅)= dE (⌅)/d⌅. An arbitrary shift of the vacuum energy the two plates, F (⌅)= dEvac(⌅)/dForce=⌅. An arbitrary vac shift of the vacuum energy by a constant that is independent of ⌅ does not aect the measurement. For the by a constant that is independentkc of=⌅ UVdoes cut noto⇤ aect the measurement. For the electromagnetic field, with two polarizations, the well-known result is that the energy density between the plates is ⇤cas = ⇥2/720⌅4. Note that this is an attractive force; electromagnetic field, with two polarizations, the well-knownvac result is that the energy as we will see, in2 the cosmic context a repulsive force requires an over-abundance of cas 1 ˙22 ⇧ 24 ˙ density betweenenergy the density: platesS is=⇤ dt= ⇤⇥ /720⇤⌅ . Note⇤ that= ⌃t⇤ this is an attractive force; vacfermions.2 2 Z ✓ ◆ as we will see, in the cosmic contextLet aus repulsive illustrate the above force remark requires on the Casimir an over-abundance eect with another version of of This effect has beenit: the measured. vacuum energy in the finite size geometry of a higher dimensional cylinder. 1 fermions. For now note: Namely, 720H == consider6 ⇧x( 120a† aa massless+ ) quantum bosonic field on a Euclidean space-time geom- 2 etry6 of S1 R3 where the circumference of the circle S1 is . Viewing the compact Let us illustrate the above remark on⇥ the Casimir eect with another version of 3 S = i⌅⌃ ⌅ i⌅⌃ ⌅ ⇧⌅⌅ it: the vacuum energy in the finite sizet geometry t of a higher dimensional cylinder. Z Namely, consider a massless quantum bosonic field on a Euclidean space-time geom- 1 etry of S1 R3 where the circumferenceH = ⇧(b†b ) of, the circleb, b† =1S1 is . Viewing the compact ⇥ 2 { }

1/3 2/3 ⇥m 3 a(t)= sinh 3 ⇥H0t/2 ⇥ ✓ ◆ h ⇣ p ⌘i a(t ) 1+z = 0 a(t)

L ⌥ = 4d2

1 d = (z + ...) H0

1 R g + g =8GT µ 2 µR µ µ

1 R g =8G (T + T ) µ 2 µR µ µ

1 Cylindrical version of Casimir eﬀect

Just change boundary conditions: join plates at edges to have periodic b.c. direction as spatial, the momenta in that direction are quantized and the vacuum One compactiﬁed energy density is spatial dimension of circumference2 β cyl 1 d k 2 2 4 3/2 ⌃ = k +(2⇧n/) = ⇧ ( 3/2)⇥( 3) + const. (2) vac 2 (2⇧)2 n ⌅ ⇤⇥ZZ ⇧ 2 dim’l space + time Due to the di⇥erent boundary conditions in the periodic verses ﬁnite size directions, cas cyl Relation to Casimir: ⌃vac(⌥)=2⌃vac( =2⌥), where the overall factor of 2 is because of the two photon

direction as spatial, the momentapolarizations. in that direction The are above quantized integral and the is vacuum divergent, however if one is only interested in its energy density is -dependence, it can be regularized using the Riemann zeta function giving the above 2 cyl 1 d k 2 2 4 3/2 ⌃ = k +(2expression.⇧n/) = Note ⇧ that( 3/ the2)⇥( constant3) + const. that(2) has been discarded in the regularization is vac 2 (2⇧)2 n ZZ ⌅ ⇤⇥ ⇧ actually at the origin of the CCP. What is measurable is the dependence. One way Due to the di⇥erent boundary conditions in the periodic verses finite size directions, cas cyl to convince oneselfdivergent that as thisUV cuto regularizationff is meaningful is to view the compactified ⌃vac(⌥)=2quantized⌃vac( =2 ⌥modes), where on the circle overall factor of 2 is because of the two photon kc → ∞. polarizations. The above integraldirection is divergent, as howeverThis Euclidean is if onethe isCosmological time, only interested where innow its =1/T is an inverse temperature. The -dependence, it can be regularized using the Riemanncylconstant zeta functionproblem. giving the above quantity ⌃vac is now the free energy density of a single scalar field, and standard expression. Note that the constant that has been discarded in the regularization is quantum7 statistical mechanics gives the convergent expression: actually at the origin of the CCP. What is measurable is the dependence. One way 3 to convince oneself that this regularization is meaningfulcyl is to1 view thed k compactified k 4 ⇥(4) ⌃ = log 1 e = . (3) vac (2⇧)3 2⇧3/2(3/2) direction as Euclidean time, where now =1/T is an inverse⌅ temperature. The cyl ⇥ quantity ⌃vac is now the free energyThe density two above of a single expressions scalar field, (2, and 3) standard are equal due to a non-trivial functional identity quantum statistical mechanics gives the convergent expression: ⇥/2 satisfied by the ⇥ function: ⌅(⇤)=⌅(1 ⇤)where⌅(⇤)=⇧ (⇤/2)⇥(⇤). (See 3 cyl 1 d k k 4 ⇥(4) ⌃ = log 1 e = . (3) vac (2⇧)3for instance the appendix2⇧3/2(3/2) in [8].) The comparison of eqns. (2,3) strongly manifests ⌅ ⇥ The two above expressions (2, 3)the are arbitrariness equal due to a non-trivial of the zero-point functional identity energy: whereas there is a divergent constant in ⇥/2 satisfied by the ⇥ function: ⌅(⇤)=⌅(1 ⇤)where⌅(⇤)=⇧ (⇤/2)⇥(⇤). (See (2), from the point of view of quantum statistical mechanics, the expression (3) is for instance the appendix in [8].) The comparison of eqns. (2,3) strongly manifests actually convergent. Either way of viewing the problem allows a shift of ⌃cyl by the arbitrariness of the zero-point energy: whereas there is a divergent constant in vac (2), from the point of view of quantuman arbitrary statistical constant mechanics, with the expression no measurable (3) is consequences. For instance, such a shift cyl actually convergent. Either waywould of viewing not thea⇥ect problem thermodynamic allows a shift of quantities⌃vac by like the entropy or density since they are an arbitrary constant with no measurablederivatives consequences. of the free For energy; instance, the such only a shift thing that is measurable is the dependence. would not a⇥ect thermodynamic quantities like the entropy or density since they are We now include gravity in the above discussion. Before stating the basic hy- derivatives of the free energy; the only thing that is measurable is the dependence. We now include gravity in thepotheses above discussion. of our study, Before we stating begin the with basic general hy- motivating remarks. All forms of energy potheses of our study, we begin with general motivating remarks. All forms of energy 4 4 TEX for keynote

1 1 1 ⇥4 1+ + + + ..... = (4) = =1.0823 24 34 44 90 TEX for keynote

1 1 1 1 z 1 1 1 ..., ⇥z4 > ( )= 1+z =1++ z++ +z +..... = (4) = ( )=11.0823 n 4 24 43 < n=1 2 3 4 90 X z 1 1 1 1 1 1t 1 (z)= (z)= =1+dt + ,+ ...,(z) > (0z) > 1 1 znz 2zt 3z < (z)(1 2n=1 ) 0 e +1 < X Z z 1 1 1 t (z)= dt , (z) > 0 ( 2) =((z)(14) =21(z) 6) ...=0et +1 < Z0

( 2)1 = ( 4) = ( 6) ...=0 z= + iy 2 1 z = + iy 3 3/2 2 d k 1 mT 2 n = = (3/2), (⇤k = k /2m) (2⇥)3 ek/T 1 2⇥ Z 3 ✓ ◆ 3/2 d k 1 mT 2 n = = (3/2), (⇤k = k /2m) (2⇥)3 ek/T 1 2⇥ Z mT ✓ ◆ n = (1) 2⇥ mT ✓ n ◆= (1) 2⇥ 1 1 1✓ ◆ (1) = 1 + + + + ...= 2 3 14 1 1 1 (1) = 1 + + + + ...= 2 3 4 1

( 3) = 1 + 23 +33 +43 + ...... =? ( 3) = 1 + 23 +33 +43 + ...... =?

1 1 = = By analytic continuation! 120 120

1 1 TEX for keynote

1 1 1 ⇥4 1+ + + + ..... = (4) = =1.0823 24 34 44 90

1 1 1 1 (z)= =1+ + + ..., (z) > 1 nz 2z 3z < n=1 X z 1 1 1 t (z)= dt , (z) > 0 (z)(1 21 z) et +1 < Z0

( 2) = ( 4) = ( 6) ...=0

1 z = + iy 2

direction as spatial, the momenta in that direction are quantized and the vacuum 3 3/2 d kenergy density1 is mT 2 n = = (3/2), (⌅k = k /2m) 3 k/T 2 cyl 1 d k 2 2 4 3/2 (2⇥) ⌃ e = 1 k2+(2⇥ ⇧n/) = ⇧ ( 3/2)⇥( 3) + const. (2) Z vac 2 (2⇧)2✓ ◆ n ⌅ ⇤⇥ZZ ⇧ QuantumDue toStatistical the di⇥erent boundary Mechanics conditions in the periodic viewpoint. verses finite size directions, cas cyl mT ⌃vac(⌥)=2⌃vac( =2n⌥),= where the overall(1) factor of 2 is because of the two photon polarizations. The above integral2⇥ is divergent, however if one is only interested in its Passing to euclidean time t = -i τ, ρvac is just✓ the finite◆ temperature free energy on the cylinder-dependence, with circumference it can be regularized β = 1/T. using the Riemann zeta function giving the above expression. Note that the1 constant1Euclidean that1 has been time discarded τ in the regularization is actually at(1) the origin = 1 of + the CCP.+ Whatwith+ is circumference measurable+ ...= is the dependence. One way 2 3β=1/T4 1 to convince oneself that this regularization is meaningful is to view the compactified direction as Euclidean time, where now =1/T is an inverse temperature. The quantity ⌃cyl is now the free energy density of a single scalar field, and standard effect withvac another version3 of it: the3 vacuum3 energy in the finite size geometry of a quantum( 3) statistical = 1 mechanics3 + dim’l 2 space+3 gives+4 the convergent+ ...... expression:=? higher dimensional cylinder. Namely, consider a massless quantum bosonic field on Quantum Statistical. Mech. 3 1 3 2 1 a Euclideancyl space-time1 d k geometry of S k R where4 the circumference⇥(4) of⇧ the4 circle S ⌃ = log 1 e = = T . (3) gives a very different vac (2⇧)3 ⊗ 2⇧3/2(3/2) 90 is β. Viewing the⌅ compact direction as spatial, the momenta in that direction are convergent expression. ⇥ The two above expressions (2, 3) are equal due to a non-trivial functional identity quantized and the vacuum energy1 density is black body = ⇥/2 satisfied by the ⇥ function:2 ⌅(⇤)=⌅(1 ⇤)where⌅(⇤)=⇧ (⇤/2)⇥(⇤). (See YES! cyl 1 d k 2 2 4 3/2 ρ = k +(2120πn/β) = β− π Γ( 3/2)ζ( 3) + const. (2) vac 2β (2π)2 − − − ? for instance then appendixZZ " in [8].) The comparison of eqns. (2,3) strongly manifests Due to the !∈ # theDue arbitrariness to the different of the boundary zero-point conditions energy: in whereas the periodic there verses is a finite divergentsize directions, constant in amazing cas cyl (2),ρ fromvac (ℓ)=2 theρ pointvac (β =2 ofz/ viewℓ),2 where of quantum the overall statistical factor of mechanics, 2 is because the of the expression two photon (3) is ⇤(z) ⇥ (z/2)(z)=⇤(1 z) functional polarizations. The above integral is divergent, however if one is only interestedcyl in actually convergent.⌘ Either way of viewing the problem allows a shift of ⌃vac by equation: an arbitraryits β-dependence, constant it with can be no regularized measurable using consequences. the Riemann For zeta instance, function giving such thea shift above expression. Note that the (infinite) constant that has been discarded in the would not a⇥ect thermodynamicAL Int. quantities J. Mod. like Phys. the entropy A23 (2008 or density) since they are regularization is actually at the origin of the CCP. What is measurable is the β derivatives of the free energy; the only thing that is measurable is the dependence. dependence. One way to convince oneself that this regularization is meaningful is to We now include gravity in the above discussion. Before stating the basic hy- view the compactified direction as Euclidean time, where now β =1/T is an inverse potheses of our study, we begincyl with general motivating remarks. All forms of energy temperature. The quantity ρvac is now the free energy density of a single scalar field, and standard quantum statistical mechanics4 gives the convergent expression which is just the standard black-body formula:

3 2 cyl 1 d k βk 4 ζ(4) π 4 ρ = log 1 e− = β− = T . (3) vac β (2π)3 − − 2π3/2Γ(3/2) −90 " $ % The two above expressions (2, 3) are equal due to a non-trivial functional identity ν/2 satisﬁed by the ζ function: ξ(ν)=ξ(1 ν)whereξ(ν)=π− Γ(ν/2)ζ(ν). (See − for instance the appendix in1 [8] in this context.) The comparison of eqns. (2,3) strongly manifests the arbitrariness of the zero-point energy: whereas there is a divergent constant in (2), from the point of view of quantum statistical mechanics, the expression (3) is actually convergent. Either way of viewing the problem allows a cyl shift of ρvac by an arbitrary constant with no measurable consequences. For instance,

4 TEX for keynote

1 1 1 ⇥4 1+ + + + ..... = (4) = =1.0823 24 34 44 90

1 1 1 1 (z)= =1+ + + ..., (z) > 1 nz 2z 3z < n=1 X z 1 1 1 t (z)= dt , (z) > 0 (z)(1 21 z) et +1 < Z0

( 2) = ( 4) = ( 6) ...=0

1 z = + iy 2

3 3/2 d k 1 mT 2 n = = (3/2), (⇧k = k /2m) (2⇥)3 ek/T 1 2⇥ Z ✓ ◆ mT n = (1) 2⇥ ✓ ◆

1 1 1 (1) = 1 + + + + ...= 2 3 4 1

( 3) = 1 + 23 +33 +43 + ...... =?

1 = 120 Riemann Hypothesis: All non-trivial zeros of z/2 Zeta have real part 1/2.⌅( z That) ⇥ is (theyz/2)( arez)= ⌅of(1 thez) ⌘ form: 1 (⇤)=0, ⇤ = + iy 1859 2

critical strip complex z-plane:

= zeros 1 critical line

functional equation: under z to 1-z, critical strip mapped onto itself x=1/2 x=1 1 1 1 (z)=1+ + + + ... 2z 3z 4z Some Riemann Zeros: 1 1 1 1 (z)= + + + ... 3.5 2z 2z 4z 6z 3.0 |ζ(1/2+i y)| 1 1 1 2.5 (1 )(z)=1+ + + ... 2z 3z 5z 2.0

1.5 1 1 1 1 (1 )(1 )(z)=1+ + + ... 1.0 3z 2z 5z 7z

0.5 1 (z)= ,p=primey 10 20 30 1 p 40z 50 p Can enumerate zero along y-axis: 1 n th zero on critical line : ⇥ = + iy n 2 n

nyn 1 14.1347251417346937904572519835624702707842571156992431756855 2 21.0220396387715549926284795938969027773343405249027817546295 3 25.0108575801456887632137909925628218186595496725579966724965 4 30.4248761258595132103118975305840913201815600237154401809621 5 32.9350615877391896906623689640749034888127156035170390092800

TABLE VI: The first few numerical solutions to (20), accurate to 60 digits (58 decimals). These 13 Known:solutions the were first obtained 10 usingzeros the are root on finder the function critical in Mathematica. line. (numerically).

279.229250927745189228409880451955359283492637405561293594727 (n =126)

Just to illustrate, and to convince the reader, how the solutions of (20) can be made arbitrarily precise, we compute the zero n = 1000 accurate up to 500 decimal places, also using the same simple approach [35]:

1419.42248094599568646598903807991681923210060106416601630469081468460 8676417593010417911343291179209987480984232260560118741397447952650637 0672508342889831518454476882525931159442394251954846877081639462563323 8145779152841855934315118793290577642799801273605240944611733704181896 2494747459675690479839876840142804973590017354741319116293486589463954 5423132081056990198071939175430299848814901931936718231264204272763589 1148784832999646735616085843651542517182417956641495352443292193649483 857772253460088 2 Substituting precise Riemann zeros calculated by other means [21] into (20) one can check that the equation is identically satisﬁed. These results corroborate that (20) is an exact equation for the Riemann zeros, which was derived on the critical line.

VIII. FINAL REMARKS

Let us summarize our main results and arguments. Throughout this paper we did not assume the Riemann hypothesis. The main outcome was the demonstration that there are

22 The distribution of Prime Numbers and Zeta

Prime number theorem How many primes less than x?

Gauss, a 15 years old boy, guessed in 1792 x ı(x)= 1 Li(x) ı log x ı pXx »

x dt Li(x)= Z0 log t

Chebyshev (1850) tried to prove using “(z) › Only proven 100 years later (1896) › by Hadamard/de la Vallé Poussin “(1 + iy) =0 6

3 TEX for keynote

1 1 1 ⇥4 1+ + + + ..... = (4)TEX = for=1 keynote.0823 24 34 44 90

4 1 1 1 1 1 1 1 ⇥ (z)= =1+1+ 4 ++4 ++4...,+ ..... = (z(4)) > =1 =1.0823 nz 2 2z 3 3z 4 < 90 n=1 X 1 z 1 1 11 t 1 1 (z)= (z)= dt=1+ , + (z+) >...,0 (z) > 1 1 z nz t 2z 3z < (z)(1 2 )n=10 e +1 < XZ z 1 1 1 t ( 2)( =z)=( 4) = ( 6)1 ...z =0dt t , (z) > 0 (z)(1 2 ) 0 e +1 < Z

1 TEXz for= keynote( +2)iy = ( 4) = ( 6) ...=0 2

4 3 1 1 1 3TEX/2 ⇥ for keynote 1+d k 4 + 41 + 4 + .....mT= (4) = =11 .0823 2 n = 2 3 4 = (390z/2)=, + iy(⇧k = k /2m) (2⇥)3 ek/T 1 2⇥ 2 Z 1✓ 1 ◆ 1 ⇥4 1+ + + + ..... = (4) = =1.0823 1 1 241 341 44 90 (z)= =1+3 + + ..., (z3)/2> 1 nz d k 2z mT31z mT< 2 nn=1= n = (1)= (3/2), (⇧k = k /2m) X (2⇥)3 ek2/T⇥ 1 2⇥ Z 1✓ 1 ◆ ✓1 1◆ (z)= =1+z 1 + + ..., (z) > 1 1 1nz t 2z 3z < (z)= n=1 dt , (z) > 0 1 z X1 1 1t mT (z(1))(1 =2 1 + ) +0 +e +1n+=...=< (1) Z z 1 2 3 1 4 12⇥1 t (z)= ✓ dt◆ , (z) > 0 (z)(1 21 z) et +1 < ( 2) = ( 4) = ( 6)...Z=00 3 3 3 1 1 1 ( 3) = 1 + 2 +3(1)+4 = 1 ++ ...... + =?+ + ...= 2 3 4 1 ( 2) = ( 4) = ( 6) ...=0 1 z = + iy 2 1 (=3) = 1 + 23 +33 +43 + ...... =? 120 1 3 3/2 z = + iy d k 1 mT 2 2 n = = (3/2), (⇧k = k /2m) 3 k/T (2⇥) e 1 z/22⇥ 1 Z ⌅(z) ⇥3 ✓ (z/◆2)(z)=⌅=(1 3/2z) ⌘d k 1 mT 2 n = = 120(3/2), (⇧k = k /2m) (2⇥)3 emTk/T 1 2⇥ Z n = (1)✓ ◆ 2⇥ 1 ✓ ◆ z/2 (⇤)=0⌅, (z)⇤ =⇥ +mTiy(z/2)(z)=⌅(1 z) Works quite well: ⌘n =2 (1) 1 1 1 2⇥ (1) = 1 + + + + ...✓= ◆ 2 3 4 1 ⇥(x) Li(x1) 1 1 1 (1) = 1( +⇤)=0+ , + ⇤ =+ ...+=iy 2 3 4 2 1 ( 3) = 1 + 23 +33 +43 + ...... =? 10 ( 3) = 1 +⇥ 23(x+3) 3 +43Li+(...... x) =? 11 = 5 120 1 = 120 z/2 1 ⌅(z) ⇥ (z/2)(z)=⌅(1 z) 10 15 ⌘20 25 30 z/2 ⌅(z) ⇥ (z/2)(z)=⌅(1 z) ⌘ 1 5 (⇤)=0, ⇤ = + iy 2 1 (⇤)=0, ⇤ = + iy 2 ⇥(x) Li(x)

⇥(x) Li(x) 150

1 100 1

200 400 600 800 1000 1 1 1 (z)=1+ + + + ... 2z 3z 4z

1 1 1 1 (z)= + + + ... 2z 2z 4z 6z

1 1 1 (1 )(z)=1+ + + ... 2z 3z 5z Zeta and the Primes 1 1 z 1 1 (1 )(1 )(z)=1+ + + ... The Golden Key: Euler 3z 2 5z 7z product formula: 1 (1737) (z)= ,p=prime 1 p z p

1 1 1 (z)=1+ + + + ... 2z 3z 4z

1 1 1 1 (z)= + + + ... 2z 2z 4z 6z

1 1 1 (1 )(z)=1+ + + ... 2z 3z 5z

1 1 1 1 (1 )(1 )(z)=1+ + + ... 3z 2z 5z 7z

1 (z)= ,p=prime 1 p z p Remark: pole at z=1 implies there are an inﬁnite number of primes.

2 the non-trivial zeros of (z). There are simpler but equivalent versions of the main result, based on the function ⇧(x) below. However, let us present the main formula for ⇤(x)itself, since it is historically more important. The function ⇤(x)isrelatedtoanothernumber-theoreticfunctionJ(x), defined as (n) J(x)= (29) log n 2 n x the non-trivial zeros of (z). There are simpler⌅ but equivalent versions of the main result, m wherebased(n on), the the function von Mangoldt⇧(x) below. function, However, is equal let us present to log p theif mainn = formulap for some for ⇤(x prime)itself,p and an integersince itm is, historically and zero otherwise. more important. The two functions ⇤(x)andJ(x) are related by Möbius inversion:The function ⇤(x)isrelatedtoanothernumber-theoreticfunctionJ(x), defined as µ(n) (n) 1/n ⇤(xJ)=(x)= J(x ). (29) (30) n log n n 12 n x Riemann’s Main ⌅Result⇥ ⌅ Here, µ(n) is the Möbius function, equal to 1 ( 1) if n is a product of an even (odd) number where (n), the von Mangoldt function, is equal to log p if n = pm for some prime p and of distinctan integer primes,m, and and zero equal otherwise. to zero The if it two has functions a multiple⇤(x)and primeJ factor.(x) are related The above by Möbius expression is actuallyinversion: a finite sum, since for large enough n, x1/n < 2andJ =0. the non-trivial zeros of (z). There are simplerµ(n) but equivalent versions of the main result, The main result of Riemann is a⇤( formulax)= for J(Jx(),x1/n expressed). as an infinite sum over(30) zeros based on the function ⇧(x) below. However, letn us present the main formula for ⇤(x)itself, n 1 ⌅ of the (z)function:since it is historically more important.⌅⇥ Here, µ(n) is the Möbius function, equal to 1 ( 1) if n is a product of an even (odd) number The function ⇤(x)isrelatedtoanothernumber-theoreticfunction J(x), defined as of distinct primes, and equal to zero if it has a multiple⇤ dt prime factor.1 The above expression J(x)=Li(x) Li (x )+ (n) log 2, (31) J(x)= log t t (t2 1) (29) x log1/nn is actually a finite sum, since for large enough2 n⇧n , xx < 2andJ=0. ⌅ ⌅ where Li(Thex)= mainwhere resultx dt/(nlog), of the Riemannt vonis the Mangoldt log-integral is a formula function, foris function equalJ(x to), logexpressed [33].p if n The= p asm abovefor an some infinite sum prime sum isp realand over because zeros the 0 ρ = a zero on the critical strip ⌅ of the (anz)function: integer m, and zero otherwise. The two functions ⇤(x)andJ(x) are related by Möbius ⌅’s come in conjugate⇤ pairs. If there are no zeros on the line (z)=1,thenthedominant inversion: ⇤ µ(n⇤) dt 1 term is the first oneJ in(x)=Li( the abovex) equation,Li⇤(x ()=x)+J(x)J(x1/nLi(). x), and thislog 2 was, used(30) to prove(31) the Derived using clever real andn complex analysis.2 n 1 x log⇥ t t (t 1) ⌅⇥ ⇧ prime numberHere, theoremµ(n) is the by Möbius Hadamard function,⌅ and equal deto 1 la ( 1) Vallée if n is a Poussin. product of an even (odd) number where Li(x)= x dt/ log t is the log-integral function [33]. The above sum is real because the The functionof distinct⇧(0x)hasthesimplerform primes, and equal to zero if it has a multiple prime factor. The above expression ⌅’s come inis conjugate actually⇤ a finite pairs. sum, If since there for large are no enough zerosn, x on1/n the< 2and lineJ =0.(z)=1,thenthedominant x 1⇤ 1 term is the firstThe main one result in the of Riemann above equation, is a formulaJ for(xJ)(x), expressedLi(x), and as an this infinite was sum used over to zeros prove the ⇧(x)= (n)=x log(2⇤) log 1 2 . (32) ⌅Remark:of the (z)function: if there are no zeros ⌅with real⇥ 2 x n x ⇥ prime number theorem⌅ by Hadamard and⌅ de la Vallée Poussin. part equal to 1, Li(x) is the leading⇤ term.dt 1 J(x)=Li(x) Li (x)+ log 2, (31) In thisThe formulation, function ⇧ the(x)hasthesimplerform prime number theorem followslog t t (t from2 1) the fact that the leading term x ⌅ ⇧ is ⇧(x) x. where Li(x)= x dt/ log t is the log-integralx function [33]. The1 above sum is1 real because the ⇥ ⇧(x)=0 (n)=x log(2⇤) log 1 . (32) ⌅ 2 x2 In Figure FIG.⌅’s come 5a in we conjugate⇤ plotn x ⇤ pairs.(x) fromIf there equations are no zeros on (30) the lineand( (31),z)=1,thenthedominant computed⇥ with the first 50 ⌅ ⌅ ⇤ term is the first one in the above equation, J(x) Li(x), and this was used to prove the zerosIn in this the formulation, approximation the prime⌅ = number1 +iy given theorem by (25).follows⇥ FIG. from 5b the shows fact that the the same leading plot with term zeros prime number theoremn by2 Hadamardn and de la Vallée Poussin. is ⇧(x) x. obtained from⇥ theThe numerical function ⇧(x)hasthesimplerform solution of equation (14). Although with the approximation yn In Figure FIG. 5a we plot ⇤(x) from⌃ equationsx (30) and1 (31), computed1 with the first 50 the curve is trying to follow⇧(x)= the steps(n)= inx ⇤(x), oncelog(2⇤ again,) log one1 clearly. sees(32) the importance ⌅ 2 x2 zeros in the approximation ⌅ n=x 1 +iy given by (25). FIG. 5b shows the⇥ same plot with zeros ⌃ of the arg term. n ⌅ 2 n ⌅ In this formulation, the prime number theorem follows from the fact that the leading term obtained from the numerical solution of equation (14). Although with the approximation yn is ⇧(x) x. ⌃ the curve is trying⇥ to follow the steps in ⇤(20x), once again, one clearly sees the importance In Figure FIG. 5a we plot ⇤(x) from equations (30) and (31), computed with the first 50 ⌃ of the arg term. 1 zeros in the approximation ⌅n = 2 +iyn given by (25). FIG. 5b shows the same plot with zeros

obtained from the numerical solution of equation (14). Although with the approximation yn ⌃ 20 the curve is trying to follow the steps in ⇤(x), once again, one clearly sees the importance ⌃ of the arg term.

20 TEX for keynote TEX for keynote

4 1 1 1 1 ⇥1 1 ⇥4 1+ 4 + 4 + 4 + ..... = 1+(4) =+ =1+ .0823+ ..... = (4) = =1.0823 2 3 4 24 9034 44 90

1 1 1 1 1 1 1 1 (z)= =1+ + (z)=+ ..., =1+(z) > 1+ + ..., (z) > 1 nz 2z 3z nz < 2z 3z < n=1 n=1 ⇤ ⇤ z 1 z 1 1 1 t 1 1(z)=t dt , (z) > 0 (z)= dt (z,)(1 2(1z)z)> 0 et +1 < (z)(1 21 z) et +1 < ⌅0 ⌅0 ( 2) = ( 4) = ( 6) ...=0 ( 2) = ( 4) = ( 6) ...=0 TEX for keynote 1 z = + iy 2 4 1 1 1 1 ⇥ z = + iy 1+ 4 + 4 + 4 + ..... = (4) = =1.0823 2 3 4 90 2 3 3/2 d k 1 mT 2 n = = (3/2), (⇧k = k /2m) 3 k/T (23/⇥2 ) e 1 2⇥ 1 3 ⌅ ⇥ 1 d k1 1 1 mT 2 (z)= n =z =1+ z + z + ...,= (z) > 1 (3/2), (⇧k = k /2m) n 23 3k/T < n=1 (2⇥) e 1 2⇥ mT ⇤ ⌅ ⇥ n = (1) 2⇥ z 1 1 1 t ⇥ (z)= dt , (mTz) > 0 (z)(1 21 z) et +1n = < (1) ⌅0 1 1 1 2⇥ (1) = 1 + + + + ...= ⇥ 2 3 4 1 ( 2) = ( 4) = ( 6) ...=01 1 1 (1) = 1 + + + + ...= ( 3) = 1 + 23 +33 +43 + ...... =? 2 3 4 1 1 z = + iy 2 1 ( 3) = 1 + 23 +33 +43 + ...... =? = 120 3 3/2 d k 1 mT 2 n = = (3/2), (⇧k = k /2m) (2⇥)3 ek/T 1 2⇥ ⌅ ⇥ z/2 1 ⌅(z) ⇥ (z/2)(z)=⌅(1 z) = ⌘ mT 120 n = (1) 2⇥ ⇥ 1 (⇤)=0, ⇤ = + iy z/2 2 1 1⌅(z1) ⇥ (z/2)(z)=⌅(1 z) (1) = 1 + + + +⌘...= 2 3 4 1 ⇥(x) Li(x) 1 8 8 (⇤)=0, ⇤ = + iy ( 3) = 1 + 23 +33 +43 + ...... =? 2 5 zeros 20 zeros 6 6 1 1 = ⇥(x) Li(x) 4 120 4

z/2 2 ⌅(z) ⇥ (z/2)(z)=⌅(1 z) 2 ⌘ 1 5 10 1 15 20 (⇤)=0, ⇤ = + iy 5 10 15 20 2

⇥(x) Li(x) 8

50 zeros 1 6

5 10 15 20 Zeta and Random Matrix Theory

The distribution of zeros on the critical line appears random, but is not completely random.

Dyson studied the properties of eigenvalues of random hamiltonians H. Though H is random, the spacing of its eigenvalues has predictable properties. (“level repulsion”)

Montgomery studied the “pair correlation function” of the zeros of zeta. Dyson pointed out that was for the same as the GUE! (1973). Veriﬁed numerically for high zeros by Odlyzko (1987)

Gaussian Unitary Ensemble = exponential of random hamiltonian 140 Random Real numbers 120

100 Eigenvalues of a 50×50 hermitian matrix 80 The ﬁrst 50 Riemann zeros 60

0.45 0.50 0.55 Substituting precise Riemann zeros into (19) one can check that the equation is identically satisﬁed. These results corroborate that (19) is an exact equation for the Riemann zeros, and we emphasize that it was derived on the critical line.

VII.referred NUMERICAL to as Dirichlet L ANALYSIS:-functions. L-FUNCTIONS There are an infinite number of distinct Dirichlet characters which are primarily char- ZetaWeacterized perform is bythe their exactly modulustrivial thek same, which case numerical determines of procedure their Dirichlet periodicity. as described They L can in- befunctions the defined previous section VID,axiomatically, but now with which equation leads to specific (42) properties, and (66) some for Dirichlet of which weL-functions, now describe. or Consider with (58) a and (67) Dirichlet character ⇥ mod k, and let the symbol (n, k) denote the greatest common divisor for L-functions based on levelDirichlet one modular characters forms. of modulus k: Axiomaticof the two definition: integers n and k. Then ⇥ has the following properties:

1. ⇥(n + k)=⇥(n). A. Dirichlet L-functions 2. ⇥(1) = 1 and ⇥(0) = 0.

We3. will⇥(nm illustrate)=⇥(n)⇥( ourm). formulas with the primitive characters 7,2 and 7,3, since they possess the full generality of a =0anda = 1 and complex components. There are actually 4. ⇥(n)=0if(n, k) > 1and⇥(n) =0if(n, k)=1. ⇥(7) = 6 distinct characters mod 7.⇥ 5. If (n, k)=1then⇥(n)⇥(k) =1,where⇤(k)istheEulertotientarithmeticfunction. Example 7,2. Consider k =7andj =2,i.e.wearecomputingtheDirichletcharacter This implies that ⇥(n)arerootsofunity. 7,2(n). For this case a =1.Thenwehavethefollowingcomponents: 6. If ⇥ is a Dirichlet character so is the complex conjugate ⇥. n 12 3 4 5 67 Example:For a given modulus k there are ⇤(k)distinctDirichletcharacters,whichessentiallyfollows (76) 2i/3 i/3 2i/3 i/3 7,2(n) 1 e e e e 10 from property 5 above. They can thus be labeled as ⇥k,j where j =1, 2,..., ⇤(k)denotesan arbitrary ordering. If k =1wehavetheTEXtrivial forcharacter keynote where ⇥(n)=1foreveryn, and The first few zeros, positive and negative, obtained by solving (42) are shown in TABLE (23) reduces to the Riemann -function. The principal character, usually denoted by ⇥1, VI (see Appendix B for the Mathematica implementation). The solutions1 shown are easily is defined as ⇥1(n)=1if(n, k)=1andzerootherwise.Intheabovenotationtheprincipal⇥ ⌅(n) ⇥ ⌅(pn) obtainedL-function: with 50 decimalL(z,⌅ places)= of accuracy,= and agree1 with the ones in [42], which were character is always ⇥k,1. nz pz n=1 n=1 n ⇥ computedCharacters up to can 20 be decimal classified places. as primitive⇤ or non-primitive⌅ . Consider the Gauss sum

k N Example 7,3. Consider k =7andj =3,suchthat2im/k a =0.Inthiscasethecomponents Also satisﬁesG(⇥)=B functional=⇥(m)e eqn.cos( . ⇥ ) (24) Nm=1 n of 7,3(n)arethefollowing: relating L(z) to L(1-zn)=1 If the character ⇥ mod k is primitive, then G(⇤⇥) 2 = k. This is no longer valid for a non- n 12| 3| 4 567 primitive character. Consider a non-primitive character ⇥ mod k. Then it can be expressed (77) 2i/3 2i/3 2i/3 2i/3 in terms of a primitive⇥ = character(tnlog) 1p of smallere arg modulus⌅e (p ) as,e⇥(nwhere)=e⇥1(n)⇥t(=n),10 where(z)⇥1 is the n 7,3 n n ⌅ principal character mod k and ⇥ is a primitive character mod k

⇥ 1 1 1 (z)= =1+ + + ..., (z) > 1 nz 2z 3z ⇤ n=1 ⇤ z 1 1 ⇥ t (z)= dt , (z) > 0 (z)(1 21 z) et +1 ⇤ ⇧0

( 2) = ( 4) = ( 6) ...=0

1 z = + iy 2

3 3/2 d k 1 mT 2 n = = (3/2), (⇧k = k /2m) (2⇤)3 ek/T 1 2⇤ ⇧ ⇥ mT n = (1) 2⇤ ⇥

1 1 1 (1) = 1 + + + + ...= 2 3 4 ⇥

( 3) = 1 + 23 +33 +43 + ...... =?

1 Strategy 1: Validity of the Euler Product

Formula to the right ofTEX critical for keynote line.

1 ⇥ ⌅(n) ⇥ ⌅(p ) L-function: L(z,⌅)= = 1 n nz pz n=1 n=1 n ⇤ ⌅ ⇥ N

BN = cos(⇥n) * Converges absolutely for Re(z) > n1=1 ⇤

* One proves there⇥ are= tnolog zerosp arg with⌅(p )Re, (wherez>1) usingt = (thez) EPF. n n n ⌅ * If the EPF is valid for Re(z)>1/2, then this combined with the functional equation impliesBN Riemann= O(⇧N) Hypothesis is true.

1 1 1 ⇤4 1+ + + + ..... = (4) = =1.0823 Conjecture 1: The Euler Product24 34 Formula44 is valid in90 the Cesaro averaged sense for Re(z)>1/2. (G.Franca and AL 1410.3520) ⇥ 1 1 1 (z)= =1+ + + ..., (z) > 1 nz 2z 3z ⇤ n=1 ⇤ The argument involved: z 1 1 ⇥ t 1. A reorganization(z)= of the series1 z for thedt tEP ,(Abel(z transform) > 0 ). (z)(1 2 ) 0 e +1 ⇤ 2. Prime number theorem, i.e. p⇧n > n log n and average gap = log n. 3. A central limit theorem for the Random Walk of Primes series ( 2) = ( 4) = ( 6) ...=0 4. The Cesaro average of the Euler product converges for Re(z)>1/2

1 z = + iy 2

3 3/2 d k 1 mT 2 n = = (3/2), (⇧k = k /2m) (2⇤)3 ek/T 1 2⇤ ⇧ ⇥ mT n = (1) 2⇤ ⇥

1 1 1 (1) = 1 + + + + ...= 2 3 4 ⇥

( 3) = 1 + 23 +33 +43 + ...... =?

1 TEX for keynote TEX for keynote TEX for keynote 1 ⇥ ⌅(n) ⇥ 1 ⌅(n) The Central⇥ L⌅((z, nLimit)⌅)=⇥ Theorem:⌅=(n) 1 L(z,⌅)= = 1nz pz nz n=1 pz1n=1 n ⇥ n=1 n=1⇤ n ⌅ ⇥ ⌅(n) ⇤⇥ ⌅⌅(n) ⇥ L(z,⌅)= = 1 N nz N pz n=1 n=1 BNn= cos(⇥n) RWPrimes series: BN = cos(⇥n) ⇥ ⇤ ⌅ n=1 n=1 ⇤ N ⇤

BN = cos(⇥n =⇥tnlog) pn arg ⌅(pn), where t = (z) ⇥n = t log pn arg ⌅(n), wheret = (z) ⌅ n=1 ⌅ ⇤ Multiplicative independence of the primes, reﬂected in their pseudo4 -randomness, 1 1 1 4 ⇤ makes the cosines behave1 1 as 1+independently1 4 + 4 + distributed4 + .....⇤ = random(4) = variables,=1.0823 1+ 4 + 4 + 4 2+ .....3= (4)4 = =1.0823 90 ⇥son =liket alog randompn walkarg2 where⌅3(pn each)4, stepwhere a randomt = 90number(z) between -1 and 1. BN = O(N) would imply convergence for Re(z)>1, ⌅but this CLT implies: ⇥ 1 1 1 ⇥ 1 (z)= 1 1=1+ + + ..., (z) > 1 (z)= =1+ +nz + ...,2z 3z(z) > 1 ⇤ nz n2=1z 3z ⇤ n=1 ⇤ ⇤ ⇧ (not fully proven yet) BN = O( N) z 1 1 z 1 ⇥ t 1(z)= ⇥ t1z dt t , (z) > 0 (z)= (z)(1dt 2 ) , (ze) >+10 ⇤ (z)(1 21 z) et +1⇧0 ⇤ ⇧0 1 The1 signiﬁcance1 of Re(⇤z4) > 1/2, i.e. right half 1+ + + + ..... = (4)( =2) = =1( 4). =0823( 6) ...=0 24 34 44 ( 2) = ( 4) = (906) ... =0 of critical strip, arises from this square root! 1 ⇥ 1 1 1 1 z = + iy (z)= =1+ + z += ...,+ iy (2z) > 1 nz 2z 3z 2 ⇤ n=1 d3k 1 mT 3/2 ⇤ 3 3/2 2 n = = (3/2), (⇧k = k /2m) d k 1 3 mTk/T 2 n = (2⇤=) e z 1 1 (3/2)2⇤, (⇧k = k /2m) (21⇤)3 ek/T⇧ ⇥1 2t⇤ ⇥ ⇧ ⇥ (z)= 1 z dt t , (z) > 0 (z)(1 2 ) 0 e +1 ⇤mT ⇧ mT n = (1) n = (1) 2⇤ 2⇤ ⇥ ⇥ ( 2) = ( 4) = ( 6) ...=01 1 1 1 (1)1 =1 1 + + + + ...= (1) = 1 + + + + ...2 =3 4 ⇥ 2 3 4 ⇥

1 ( 3) = 1 + 23 +33 +43 + ...... =? (z =3) = 1+ +iy 23 +33 +43 + ...... =? 2 1 1 = 3 3/2 = 120 d k 1 mT 120 2 n = = (3/2), (⇧k = k /2m) (2⇤)3 ek/T 1 2⇤ ⇧ ⇥ 1 mT 1 n = (1) 2⇤ ⇥

1 1 1 (1) = 1 + + + + ...= 2 3 4 ⇥

( 3) = 1 + 23 +33 +43 + ...... =?

1 Numerical Evidence is compelling

BN BN

300 600

200 400

100 200

N N 104 5104 105 1105 2105 3105 4105 5105

FIG. 1. The absolute value of the partial sum BN versus N, for a ﬁxed t. Left: We use (23) with t =5 103. Note that N is below the cut-o (30). Right: Here we use (21) (u = 1) with the · character = shown in (A3), and t =5 102. In this case we can freely take the limit N . 7,2 · ⇥⇤

For non-principal Dirichlet L-functions, the analog of the analytic continuation (31) con- tains log L(ns, ⇤)insteadoflog(ns). Since L(s, ⇤)hasnopole,allthepreviousdiscussion

does not apply. In this case2 x there is no need for a cut-o Nc, and we2 canx take N N0,⇥ N0,⇥ ⇥⇤ 0.7 regardless of t. 0.6 2.0 ⇥ ⇥ 0.5 1.5 0.4

D. Numerical veriﬁcation0.3 1.0

0.2 0.5 The convergence of B0.1N /⌅N then does not rely on any special detailed properties of the x x primes, but rather the opposite, on1 their multiplicative0 1 independence.2 Let us numerically1 0 1 2

test our previous conclusionsFIG. 2. Left: for (23),The related probability to the density-function. function Obviously, of RN /⇧N the(red same dashed is true line) compared with the for non-principal DirichletdensityL of-functionsB /⇧N through(solid blue (21). line). In Wethis used caset the= 10 numerical3, N =3 evidence104 and isE =8 104 ensembles. N · · even better, and we doNumerically not need a these cut-o distributions.Letuscheckthegrowthfor(23)predictedbythe are a gaussian with variance 2 0.578 and approximately zero ⇤ previous argument. Inmean. FigureRight: 1 weexactly plot the the partial same sums numericalBN experimentand one clearly but with sees the this replacement⌅N p n log n in | | n ⇤ growth. Note that we choose a range N

Let us also conﬁrmdistribution. the gaussian In Figure distribution 2 (left) as we stated plot in the (17), this using density the for additional both RN /⇧N and BN /⇧N,

freedom that comes fromwhere theR randomN is given variable by (16)u. withIt is importantrn arandomvariableon[ to note that u is not1, 1]. chosen One sees that they are independently for eachnearly cosine indistinguishable. term in the sum; This rather is compelling one chooses numerical a ﬁxed u, evidence randomly, that thenBN /⇧N approaches a ⇧ computes the sum BgaussianN (u). In distribution this sense at a singlelarge N sum, withBN zero(u) mean is completely and variance deterministic 1/2, since RN / N obeys the 1 u2 for a given u. NowCLT consider with a an normal ensemble distributionB (u )/⌅eN .E In, Figure where 2 for (right) each we element can also of see that if we use the { N i ⇥ }i=1 the set we choose aPNT random andu replacein thepn intervaln log [0n,in2⇥]. (21), Now we we loose can this consider normal its distribution. density This is expected i ⇤ since with this approximation the n’s will not be linear independent. This shows that the 14 CLT for (21), or (23), comes from ﬂuctuations in the primes, which are not captured by a smooth approximation.

IV. L-FUNCTIONS BASED ON MODULAR FORMS

Another important class of L-functions that also enjoy a functional equation and Euler product formula are based on modular forms of weight k, where k 4isaneveninteger. ⇥ Hecke theory explains how the nice properties of the L-functions, analyticity, functional equation, and Euler product formula, arise from the modularity. The functional equation relates L(s)toL(k s)thusthecriticallineis⇥ = (s)=k/2. In this section we brieﬂy ⌅ describe how to extend the above results to this class of L-functions. For simplicity we consider the example of k =12,sinceitisawell-knownfunction,anditwillbeclearhow

15 1. Riemann -function

In Figure 4 one can see how the partial product in (A1) converges to the (s)function as we increase N. For higher N the curves are indistinguishable. Note, however, that we cannot go beyond the cut-o (30) for a given t. Let us also verify convergence for arg , which plays a central role for the zeros on the critical line (see the Section V). Using the EPF we have equation (42), whose equality is veriﬁed in Figure 5. This assures that both the real and imaginary parts of the Euler product converge. As we approach the critical line ⇥ 1/2+ higher N is of course required. ⇥ One can clearly see how the Euler product formula is not valid for ⇥ 1/2 from Figure 6. The curves only match for ⇥ > 1/2andthedramaticchangeinbehaviorisabruptat⇥ =1/2, as predicted. The divergences shown in Figure 6 (right) get worse for higher N. As we discussed before, there is no convergence on the real line t =0duetothepole S⇥ t at s =1.Itisexactlybecauseofthisdivergencethatwehadtointroducethecut-o0.6 (30). However, for short truncations of the product, i.e. not so high N, we can describe ⇥ the -function quite accurately even for low t, as shown in Figure 7 (left). We can see that 0 the oscillations get stronger close to t =0,buttheCes`aroaverageisstillwell-behavedand ⇤ 3 4it ⇤ 3 4it

5 N⇥5 5 N⇥100 0.6 ⌅ ⇤ ⇥⌅ ⌅ ⇤ ⇥⌅ 4 4 t 1005 1010 1015

3 3

1 1 2 2 FIG. 5. The black line is the actual arg ⇥ + + it , and the blue line is the RHS of (42). Dots 2 1 1 1 5 ⇥ are added to the line to aid in visualization. We used = 10 and N = 10 . t t 1000 1005 1010 1015 1000 1005 1010 1015 ⇤ 3 4it ⇤ 3 4it closer to the actual value of (s)thanthepartialproductitself.Theconvergencecloseto 5 N⇥1000 5 N⇥10000 the⌅ critical ⇤ ⇥⌅ line is very slow, and requires very high⌅ ⇤ N⇥⌅. Thus to test the results close to 4 4 the critical3 line, we also have to choose high t due to3 the cut-o relation (30). One can see from Figure2 7 (right) that the Ces`aro average approximates2 (s)correctly,evenclosetothe 1 1 critical line. t t In Table1000 I we1005 show some1010 values of1015 the average 1000 and1005 the product1010 itself.1015 The |⇥PN ⇤| |PN | FIG.convergence 4. The black is slow, line is but the one actual can see(3/4+ thatit) , analytically(s) ( continueds)asweincrease into the strip,N, whereasand the blue the | |⇥PN ⇤ lineunaveraged is the partial( products)continuestooscillatearoundN (3/4+it) . Dots are added(s). to With the lineN to=10 aid5 visualization.we obtain nearly 5 PN |P | 23 ⇤ ⇥it

15 ⇤ 15 ⇥⇤

10 10

5 5

⇥ 0.5 1 t 1000 1005 1010 1015 FIG. 6. Left: the black line corresponds to ⇥(⇤ + it) against 0 < ⇤ < 1, for t = 500. The blue | | line is the partial product (⇤ + it) with N = 104. Right: the black line is the exact ⇥ , and |PN | | | the blue line is the partial product (with N =8 103), against t. We took ⇤ =0.4. The red |PN | · dots are the Ces`aro average . If we increase N the results are even worse. |⇥PN ⇤|

24 ⇥ 0.8it ⇥ 0.55it 3.0

8 ⇤ 2.5 ⇥⇤ ⇤ ⇥⇤

2.0 6

1.5 4 1.0 2 0.5 t t 5 10 15 20 1031 10310

FIG. 7. Left: the black line is (0.8+it) , and the blue line is ,withN = 102, and the | | |PN | red dots correspond to the Ces`aro average . Right: here we have (0.55 + it), and we used |⇤PN ⌅| N =4 104. The convergence close to the critical line requires very large N. ·

digit accuracy for t = 100. Note that the results eventually start to get worse for very high N, here roughly 108. We are increasing N much beyond the cut-o predicted by (30). We can do this since we are not close to the critical line.

N N |⇤PN ⌅||PN | |⇤PN ⌅||PN | 1 103 0.976752 0.972210 1 103 1.690988 1.694894 · · 2 103 0.976690 0.981506 2 103 1.692350 1.694156 · · 3 103 0.977653 0.976654 3 103 1.692590 1.690354 · · 4 103 0.977865 0.975735 4 103 1.692399 1.688480 · · 5 103 0.977926 0.984674 5 103 1.691996 1.687150 · · 6 103 0.977463 0.977893 6 103 1.691666 1.689158 · · 7 103 0.978208 0.976510 7 103 1.691508 1.688145 · · 8 103 0.977593 0.978773 8 103 1.691400 1.691700 · · 9 103 0.978290 0.981781 9 103 1.691381 1.692973 · · 1 104 0.977900 0.971017 1 104 1.691345 1.690480 · · 1 105 0.977703 0.971203 1 105 1.691373 1.692136 · · 1 106 0.977925 0.971491 1 106 1.691429 1.691577 · · 1 107 0.978168 0.978027 1 107 1.691414 1.691703 · · 1 108 0.977823 0.984481 1 108 1.691385 1.693287 · · 2 108 0.956304 0.885545 2 108 1.745257 1.923738 · · 3 108 0.924928 0.794254 3 108 1.852499 2.203470 · · (0.95 + i 20) =0.977848 (0.95 + i 100) =1.691397 | | | | TABLE I. Convergence , and , for the -function. Note that even for N N t2 the ⇤PN ⌅ PN ⇥ c results are good, but eventually it starts to deviate from the correct value as shown in the two last entries.

25 Zeta and other L’s based on principal characters are the exception and actually trickier since all characters are 1 or 0. Partly due to the pole at z=1. Transcendental equations for individual zeros.

AL Int. J. Mod. Phys. A28 (2013) G. França, AL, Comm. Numb. Theory and Phys. 2015 Everyone here knows one function with an cos(z) =0 inﬁnite number of zeros along a line in the complex z-plane...... for z=(n+1/2)π

Our result: There are an inﬁnite number of zeros of zeta along critical line in one-to-one correspondence with the zeros of cosine.

The n-th zero satisﬁes a Transcendental Equation that depends only on n. 1 1 1 (z)=1+ + + + ... 2z 3z 4z 1 1 1 (z)=1+ + + + ... 2z 3z 4z 1 1 1 1 (z)= + + + ... 2z 2z 4z 6z 1 1 1 1 (z)= + + + ... 2z 2z 4z 6z 1 1 1 (1 )(z)=1+ + + ... 2z 3z 5z 1 1 1 (1 )(z)=1+ + + ... 2z 3z 5z 1 1 1 1 (1 )(1 )(z)=1+ + + ... 3z 2z 5z 7z 1 1 1 1 (1 )(1 )(z)=1+ + + ... 3z 2z 5z 7z 1 (z)= ,p1 =prime1 1 1 p(z)=1+ + + + ... 1 p 2z 3z 4z (z)= ,p=prime 1 p z p 1 1 11 1 1 (z)=1+ + + + ...1 (z)=z z + z + + ... n th zero on critical2z line2 : 23z ⇤4z = 6z + iy n 2 n 1 1 1 1 1 n th zero on critical line :(z)=⇤n1 =+ ++ iyn+1 ... 1 2z (1 2)z(z2)=1+4z 6z + + ... 2z 3z 5z ((n +1)=n!) 1 1 1 (1 1)(z)=1+1 + + ...1 1 (1 2z )(1 )(3zz)=1+5z + + ... ((n +1)=n!)3z 2z 5z 7z 1 1 1 1 (1 )(1 )(z)=1+ + + ... ⇥z (1 +z iy) 1 z z 3 (z)=2 ,p5 7 =prime 1 p z p ⇥(1 + iy) 1 (z)= ,p=prime 1 p z 1 p n th zero on critical line : ⇤n = + iyn ⇤1 ⇤2 2 1 n th zero on critical line : ⇤n = + iyn ⇤1 ⇤2 2 ((n +1)=n!) i(x,y) ⌅(z)=⌅(x + iy)=((n A+1)=(x, yn)!)e ⌅(z)=⌅(x + iy)=A(x, y)ei(x,y⇥)(1 + iy) ⇥(1 + iy) i (x,y) ⌅(1 z)=⌅(1 x iy)=A⇥(x, y)e 0 ⇤1 ⇤2 i0(x,y) How⌅ to(1 derivez)= this⌅(1 equation:x iy )=A⇥(x,⇤1 y)e ⇤2 ⌅(z)=⌅(x + iy)=A(x, y)ei(x,y) A⇥(x, y)=A(1 x, y), ⇥⇥(x, y)=⇥i(1(x,y) x, y) Write: ⌅(z)=⌅(x + iy)=A(x, y)e

A⇥(x, y)=A(1 x, y), ⇥⇥(x, y)=⇥(1 x, y) i (x,y) ⌅(1 z)=⌅(1 x iy)=A⇥(x, y)e 0 i (x,y) ⌅(1 z)=⌅(1 x iy)=A⇥(x, y)e 0 If ⇤ is a zero : ⌅(⇤)+ ⌅(1 ⇤)=0 If ⇤ is a zero : ⌅(A⇤⇥)+(x, y)=⌅(1A(1 ⇤)=0x, y), ⇥⇥(x, y)=⇥(1 x, y) A⇥(x, y)=A(1 x, y), ⇥⇥(x, y)=⇥(1 x, y) i i = e + e 0 =0 ⇥

2 The particular solution2 = , cos()=0 givesan inﬁnitenumberofzerosonthecriticalline 2 The particular solution = , cos()=0 givesan inﬁnitenumberofzerosonthecriticalline2

There exists an inﬁnite number of zeros of zeta satisfying

θ(x=1/2, y) = (n+1/2) π

3 where i⇥(x,y) i⇥ (x,y) B(x, y) e + e 0 . (11) ⇤ + The second equality in (10) follows from A = A⇤. Then, in the limit 0 , a zero ⇧ corresponds to A =0,B =0orboth.Theycansimultaneouslybezerosincetheyare not independent. If B =0thenA =0,sinceA ⇥(z) . However, the converse is not ⌃ | | necessarily true. Since there(a) is more structure in B, let us consider B =0.Thegeneralsolutionofthis(b)

equation is given by ⇤+1⇤⇤ =(2n1+1)⌅, which are a family of curves y(x). However, since ⌃(z) FIG. 2: (a) A plot of S(y)= arg 2 + iy as a function of y showing its rapid oscillation. The is an analytic function, we know that the zeros must be isolated points rather than curves, jumps occur on a Riemann zero. (b) The function⇥ N (T ) in (15), which is indistinguishable from and this general solution must be restricted. Thus,0 let us choose the particular solution a manual counting of zeros. ⇤ = ⇤⇤, lim cos ⇤ =0. (12) 0+ ⇥ III. APPROXIMATE SOLUTION IN TERMS OF THE LAMBERT FUNCTION On the critical line, the first equation (12) is already satisfied. Now, in the limit 0+, ⇧ 1 the second equation implies ⇤ = n + 2 ⌅, for n =0, 1, 2,..., hence A. Main formula ± ± y y ⇥ 5 1 n = log +lim arg ⇥ 1 + iy + . (13) 2⌅ 2⌅e 8 0+ ⌅ 2 Conjecture 2: θ(x=1/2, y) = (n+1/2) π is⇥ a Let us now show that if one neglects⇤ ⌅ the arg term, the equation⇥ (14) can be exactly A closertranscendental inspection shows equation that the for rightthe ordinate hand side yn ofof (13)the n has-th a minimum in the interval solved. First, letRiemann us introduce zero: the Lambert W function [25], which is defined for any complex The particular solution( 2, 1), = thus, n cos(is bounded)=0 from givesan below, infinitenumberofzerosonthecriticalline i.e. n 1. Establishing the convention that ⌅ number z through the equation 1 zerosThere are labeled is an exact by positiveequation, integers, but let ⇧men = present2 + iy nitswhere large yn limit:=1, 2,..., we must replace W (z)eW (z) = z. (23) n n 2 in (13). Therefore, the imaginary1 parts of these zeros are determined from the ⇧ The n th zero is of the form ⇥ = + iy 2 n The multi-valuedsolution of theW transcendentalfunction cannot equation be expressed in terms of other known elementary

yn yn 1 1 11 functions. If we restrictlog attention+lim to real-valuedarg ⇥ + W+(iyxn)therearetwobranches.Theprincipal= n (n =1, 2,...). (14) 2⌅ 2⌅e 0+ ⌅ 2 8 The particular⇥ solution = , cos()=0 givesan infinitenumberofzerosonthecriticalline branch occurs when W⇤(x) ⌅ 1andisdenotedby ⇥ W0, or simply W for short, and its In short, we have shown⇤ that, asymptotically, there are an infinite number of zeros on the domain is x 1/e. The secondary branch, denoted by W 1, satisfies W 1(x) 1for critical⇤ line whose ordinates can be determined by solving (14) for yn. ⇥ 1 1 e x

3 n yn 1022 +1 1.370919909931995308226770 1021 ⇧ ⇥ 1050 5.741532903784313725642221053588442131126693322343461 1048 ⇥ 10100 2.80690383842894069903195445838256400084548030162846045192360059224930 922349073043060335653109252473234 1098 ⇥ 200 10 (a) 1.38579222214678934084546680546715919012340245153870708183286835248393(b) 8909689796343076797639408172610028651791994879400728026863298840958091 FIG. 2: (a) A plot of S(y)= 1 arg288304951600695814960962282888090054696215023267048447330585768 1 + iy as a function of y showing its rapid oscillation. The 10198 2 ⇥ jumps occur on aTABLE Riemann I: Formulazero. (b) (25) The can function⇥ easily estimateN0(T ) very in (15), high which Riemann is indistinguishable zeros. The results fromare expected to a manual countingbe of correct zeros. up to the decimal point, i.e. to the number of digits in the integer part. The numbers are shown with three digits beyond the integer part. III. APPROXIMATE SOLUTION IN TERMS OF THE LAMBERT FUNCTION The solution iswe explicitly obtain 11 given in terms of an 2⇥ n 8 A. Main formula yn = . (25) elementary function: the W e 1 n 11 ⇥8 Lambert W-function: Let us now showAlthough that if one the inversion neglects from the arg (24)⇧ toterm, (25)⇤ the is rather equation⇥⌅ simple, (14) it is can very be convenient exactly since it is indeed an explicit formula depending only on n,andW is included in most numerical solved. First, let us introduce the Lambert W function [25], which is defined for any complex packages. It gives an approximate solution for the ordinates of the Riemann zeros in closed number z through the equation form. The values computed from (25) are much closer to the Riemann zeros than Gram W is defined to satisfy: W (z)eW (z) = z. (23) points, and one does not have to deal with violations of Gram’s law (see below). The multi-valued W function cannot be expressed in terms of other known elementary functions.Lambert If we restrictW B.was attention Furtherfirst studied remarks to real-valued by LambertW (inx )therearetwobranches.Theprincipalthe 1758. Euler recognized its importance in 1779 in branch occurs when W (x) 1andisdenotedbyW0, or simply W for short, and its a paper onRemark transcendental7. The⇤ estimates equations, given and by (25)credited can be calculated to high accuracy for arbitrarily domain is x 1/e. The secondary branch, denoted by W 1, satisfies W 1(x) 1for Lambert.⇤ ⇥ 1 large n, since W is a standard elementary function. Of course, the yn are not as accurate e x<0. Since we are interested in positive real-valued solutions of (14), we just need ⇥ as the solutions yn including the arg term, as we will see in section IV. Nevertheless, the principalIt’s importance branch where was onlyW is realized single-valued. in the 1990’s, ⇧ when it finallyit is indeed obtained a good the estimate, name especially if one considers very high zeros, where traditional Let us consider the leading order approximation of (14), or equivalently, its average since the Lambertmethods W-function. have not previously estimated such high values. For instance, formula (25) can arg 1 + iy =0.Thenwehavethetranscendentalequation ⌅ 2 ⇧ easily estimate the zeros shown in TABLE I, and much higher if desirable. The numbers in ⇥ this table are accuratey approximationsy to the11 n-th zero to the number of digits shown, which n log n = n . (24) 2⇥ 2⇥e 8 is approximately the number⇧ of⌃ digits in the integer part. For instance, the approximation 100 ⌥ ⌥ to the 10 zero is correct to11 100 digits.1 With Mathematica we easily calculatedxn the first Through the transformation yn =2⇥ n 8 xn , this equation can be written as xne = 106 1 11 million digits of the 10 zero. 1 11 e n 8 . Comparing with (23) its solution⇥ is given by xn = W e n 8 ,andthus ⌥ ⇥ ⇤ ⇥⌅ 13 12 22,048-UWO Poster 9/1/04 3:31 PM Page 1

A Fractal Related to Johann Heinrich Lambert Each colour represents a cycle length in the iteration Johann Heinrich Lambert was born in Mulhouse on the 26th of August, 1728, and died in , with . A pixel at coordinate Berlin on the 25th of September, 1777. His scientific interests were remarkably broad. The where is given the colour corresponding to the length of the attracting cycle. self-educated son of a tailor, he produced fundamentally important work in number theory, geometry, statistics, astronomy, meteorology, hygrometry, pyrometry, optics, cosmology and philosophy. Lambert was the first to prove the irrationality of . He worked on the parallel postulate, and also introduced the modern notation for the hyperbolic functions. In a paper entitled “Observationes Variae in Mathesin Puram”, published in 1758 in Acta Helvetica, he gave a series solution of the trinomial equation, , for His method was a precursor of the more general Lagrange inversion theorem. This solution intrigued his contemporary, Euler, and led to the discovery of the Lambert function. Lambert wrote Euler a cordial letter on the 18th of October, 1771, expressing his hope that Euler would regain his sight after an operation; he explains in this letter how his trinomial method extends to series reversion. The Lambert function is implicitly elementary. That is, it is implicitly defined by an equation containing only elementary functions. The Lambert function is not, itself, an elementary function. It is also not a Liouvillian function, which means that it is not expressible as a finite sequence of exponentiations, root extractions, or antidifferentiations (quadratures) of any elementary function. The Lambert function has been applied to solve problems in the analysis of algorithms, the spread of disease, quantum physics, ideal diodes and transistors, black holes, the kinetics of pigment regeneration in the human eye, dynamical systems containing delays, and in many other areas.

The graph of for real values of and Leonhard Euler Leonhard Euler was born on the 15th of April, 1707, in Basel, Switzerland, and died on the 18th of September, 1783, in St. Petersburg, Russia. Half his papers were written in the last fourteen years of his life, even though he had gone blind. Euler was the greatest mathematician of the 18th century, and one of the greatest of all time. His work on the calculus of variations has been called “the most beautiful book ever written”, and Pierre Simon de Laplace exhorted his students: “Lisez Euler, c’est notre maître â tous”, advice that is still profitable today.

Equipotentials and electric field Many functions and concepts are named after him, including the Images of circles and rays under the maps lines at the edge of a capacitor Euler totient function, Eulerian numbers, the Euler-Lagrange equations, and the “eulerian” formulation of fluid mechanics. The . Equivalently, images of consisting of two charged thin mathematical formulae on this poster are typeset in the Euler font, designed by Hermann Zapf to evoke horizontal and vertical lines under the map plates a distance apart. the flavour of excellent human handwriting. . Lambert’s series solution of his trinomial equation, which Euler rewrote as , led to the series solution of the transcendental equation . This was the earliest known occurrence of the series for the function now called the Lambert function.

A portion of the Riemann surface for , drawn by plotting a surface with height Hippias of Elis at coordinates and Hippias of Elis lived, travelled and worked around 460 BC, and is colouring the surface with Re ; mentioned by Plato. The Quadratrix (or trisectrix) of Hippias is the first the apparent intersection on the line curve ever named after its inventor. As drawn in the picture here, its is of surfaces equation is . This curve can be used to square the circle and to trisect the angle. Since these classical problems are unsolvable by with different colours and therefore not a straightedge and compass, we therefore conclude that the construction true intersection. of the Quadratrix is impossible under that restriction. The Quadratrix is also the image of the real axis under the map , and the parts of the curve corresponding to the negative real axis delimit the ranges of the branches of . We have here coloured the ranges of the different branches of with different colours.

Sir Edward Maitland Wright Sir Edward Maitland Wright was born the 1st of January, 1906. He is the co-author with G. H. Hardy of the classic book An Introduction to the Theory of Numbers. His main contributions to the study of the R.M. Corless, G.H. Gonnet, D.E.G. Hare, D.J. Jeffrey, and D.E. Knuth, Lambert function were a systematic way of computing its complex “On the Lambert Function”, Advances in Computational Mathematics, volume 5, 1996, pp. 329-359 values, a series expansion of a related function about its branch points, the application of to enumeration problems, and the application of to the study of the stability of the solutions of linear and nonlinear delay differential equations. He was Professor of Mathematics, then Principal www.orcca.on.ca/LambertW and Vice-Chancellor, of Aberdeen University (1936-1976). n yn yn 1 14.52 14.134725142 With Lambert W one can accurately 10 50.23 49.773832478 2 estimate arbitrarily high zeros, even 10 235.99 236.524229666 103 1419.52 1419.422480946 1000000 the 10 -th to million digit 104 9877.63 9877.782654006 accuracy. 105 74920.89 74920.827498994 106 600269.64 600269.677012445 107 4992381.11 4992381.014003180 108 42653549.77 42653549.760951554 109 371870204.05 371870203.837028053 1010 3293531632.26 3293531632.397136704 The 10999 -th zero to 1000 digits TABLEbased II: Numerical on Lambert solutions of equation (14). All numbers shown are accurate up to the 9-th decimal place, in comparison with [21, 27].

W: 2.7418985289770733523380199967281384304396404342236129703462008148794017483102 288989728527567413645122744311921172826961083680270092169498827568635959416113nyn 429885386834142256620793027203450326850405406192401605278151278292126757823589 1 14.13472514173469379045725198356247 021159380557496232240667437943583994705834760582066723674368091278444158666608 4559778530181772820265652672552738836014990753552174441892311047526844245934382 21.02203963877155499262847959389690 624806198537729334547336147304637269663107947384735659921127394121662743671648 996 2112948866018589452794962947279550946390292880940546879412522254784267861820463 25.01085758014568876321379099256282 × 10 5232217042630950851351008193833985961697039872283360440246593500887533853245374 30.42487612585951321031189753058409 829732202404696954235778305250096210562727012320495894109605623304319565563992 4847173806377094362402204521511110449393462819512496547469875401348247138713215 32.93506158773918969066236896407490 328533373296657458895502274291514524646315414320664466625774466094199153901000 163674331154397634011868264241305320165870441692798635788965590575893640872077TABLE III: Numerical solutions to (14) for the lowest zeros. Although it was derived for high y, 63792090920744162661827244311481936682248189296258020149248439142 Diﬀer only in it provides accurate numbers even for the lower zeros. This numbers are correct up to the decimal last digit shown The 10999 +1 -th zero to 1000 digits basedplace shown on [21]. to 9 digits after the integer part. Lambert W: 2.7418985289770733523380199967281384304396404342236129703462008148794017483102 288989728527567413645122744311921172826961083680270092169498827568635959416113Although the formula for yn was derived for large y, it is surprisingly accurate even for 429885386834142256620793027203450326850405406192401605278151278292126757823589the lower zeros, as shown in TABLE III. It is actually easier to solve numerically for low 021159380557496232240667437943583994705834760582066723674368091278444158666608 455977853018177282026565267255273883601499075355217444189231104752684424593438zeros since arg is better behaved. These numbers are correct up to the number of digits 624806198537729334547336147304637269663107947384735659921127394121662743671648 shown, and the precision was improved simply by decreasing the error tolerance. 211294886601858945279496294727955094639029288094054687941252225478426786182046 × 10996 523221704263095085135100819383398596169703987228336044024659350088753385324537Riemann zeros have previously been calculated to high accuracy using sophisticated algo- 829732202404696954235778305250096210562727012320495894109605623304319565563992 484717380637709436240220452151111044939346281951249654746987540134824713871321 32853337329665745889550227429151452464631541432066446662577446609419915390100016 163674331154397634011868264241305320165870441692798635788965590575893640872077 63792090920744162661827244311481936682248189296258020149248439145 Solutions of the asymptotic transcendental equation are accurate enough to reveal the GUE statistics:

M = 109 105,N= 109

1.0

5 10 zeros around the 0.8 billion-th zero: 0.6 curve is the GUE prediction 0.4

0.2

0.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 (a) (b)

FIG. 4: (a) The solid line represents theLambert RHS approximation of (28) and not the good dots enough represent to see the its statistics. LHS, computed from the numerical solutions of equation (14). The parameters are ⇥ = +0.05, =[0, 0.05,...,3]

1 5 and the x-axis is given by x = 2 ( + ⇥). We use the ﬁrst 10 zeros. (b) The same parameters but using zeros in the middle of the critical line, i.e. M = 109 105 and N = 109. nyn nyn 105 5 74917.719415828 109 5 371870202.244870424 105 4 74918.370580227 109 4 371870202.673284411 105 3 74918.691433454 109 3 371870203.177729726 105 2 74919.075161121 109 2 371870203.274345875 105 1 74920.259793259 109 1 371870203.802552402 105 74920.827498994 109 371870203.837028146

TABLE V: Last numerical solutions to (14) around n = 105 and n = 109. In the ﬁrst table the solutions are accurate up to the 8-th decimal place, while in the second table up to the 6-th decimal place [21, 27]. statistics. In fact, FIG. 4a is identical to the one in [30]. The last zeros in these ranges are shown in TABLE V.

VI. PRIME NUMBER COUNTING FUNCTION

In this section we explore whether our approximations to the Riemann zeros are accurate enough to reconstruct the prime number counting function. As usual, let (x)denotethe number of primes less than x. Riemann obtained an explicit expression for (x)intermsof

19 nyn 1 14.1347251417346937904572519835624702707842571156992431756855 2 21.0220396387715549926284795938969027773343405249027817546295 3 25.0108575801456887632137909925628218186595496725579966724965 4 30.4248761258595132103118975305840913201815600237154401809621 5 32.9350615877391896906623689640749034888127156035170390092800

TABLE VI: The ﬁrst few numerical solutions to (20), accurate to 60 digits (58 decimals). These solutions were obtained using the root ﬁnder function in Mathematica.

It is known that the ﬁrst zero where Gram’s law fails is for n = 126. Applying the same method, like for any other n, the solution of (20) starting with the approximation (25) does not present any di⇥culty. We easily found the following number: Solving the exact version of the transcendental 279.229250927745189228409880451955359283492637405561293594727 (n =126) equation gives zeros to any desired accuracy. Just to illustrate, and to convince the reader, how the solutions of (20) can be made arbitrarily precise, we compute the zero n = 1000 accurate up to 500 decimal places, also using Thethe 1000 same-th simple zero to approach 500 digits: [35]:

Substituting precise Riemann zeros calculated by other means [21] into (20) one can check ...... with very simple Mathematica that the equation is identically satisﬁed. These results corroborate that (20) is an exact commands. equation for the Riemann zeros, which was derived on the critical line.

VIII. FINAL REMARKS

Let us summarize our main results and arguments. Throughout this paper we did not assume the Riemann hypothesis. The main outcome was the demonstration that there are

22 where i⇥(x,y) i⇥ (x,y) B(x, y) e + e 0 . (11) ⇤ + The second equality in (10) follows from A = A⇤. Then, in the limit 0 , a zero ⇧ corresponds to A =0,B =0orboth.Theycansimultaneouslybezerosincetheyare not independent. If B =0thenA =0,sinceA ⇥(z) . However, the converse is not ⌃ | | necessarily true. Since there is more structure in B, let us consider B =0.Thegeneralsolutionofthis

equation is given by ⇤+⇤⇤ =(2n+1)⌅, which are a family of curves y(x). However, since ⌃(z) is an analytic function, we know that the zeros must be isolated points rather than curves, and this general solution must be restricted. Thus, let us choose the particular solution

⇤ = ⇤⇤, lim cos ⇤ =0. (12) 0+ ⇥ On the critical line, the ﬁrst equation (12) is already satisﬁed. Now, in the limit 0+, ⇧ the second equation implies ⇤ = n + 1 ⌅, for n =0, 1, 2,..., hence 2 ± ± y y ⇥ 5 1 n = log +lim arg ⇥ 1 + iy + . (13) 2⌅ 2⌅e 8 0+ ⌅ 2 ⇥ ⇤ ⌅ ⇥ A closer inspection shows that the right hand side of (13) has a minimum in the interval

The particular solution( 2, 1), thus= , n cos(is bounded)=0 from givesan below, inﬁnitenumberofzerosonthecriticalline i.e. n 1. Establishing the convention that Strategy 2 to prove the Riemann⌅ Hypothesis 1 zeros are labeled by positive integers, ⇧n = 2 + iyn where n =1, 2,..., we must replace n n 2 in (13). Therefore, the imaginary1 parts of these zeros are determined from the ⇧ The n th zero is of the form ⇥ = + iyn Recall our solution of the transcendental equation 2 main result: yn yn 1 1 11 log +lim arg ⇥ + + iyn = n (n =1, 2,...). (14) 2⌅ 2⌅e 0+ ⌅ 2 8 ⇥ ⇤ ⌅ ⇥ In short, we have shown that, asymptotically, there are an inﬁnite number of zeros on the If there is a unique solutioncritical to line this whose equation ordinates for can every be determined n, since by they solving are (14) enumerated for yn. by n, we can count howNote many that, byzeros comparing are on with the the critical counting functionline upN to(T ), a theheight left hand y=T. side of (14) is amonotonicincreasingfunctionofy, and the leading term is a smooth function. Possible discontinuities can only come from 1 arg ⇥ 1 + iy , and in fact, it has a jump discontinuity N0(T) = number of zeros on the line with ordinate⇤ 2y

additional zeros from A =0in(10)norfromthegeneralsolution⇤ + ⇤⇤ =(2n +1)⌅. This strongly suggests that (14) describes all the non-trivial zeros, which are all on the critical line.

B. Exact equation 3

Let us now reproduce the same analysis discussed previously but without an asymptotic expansion. The exact versions of (6) and (7) are

x/2 1 A(x, y)=⌅ (x + iy) ⇥(x + iy) , (16) | 2 || | y ⇤(x, y)=arg 1 (x + iy) ⇥ log ⌅ +arg⇥(x + iy), (17) 2 2 i⇥ ⇥i⇥ where again ⌃(z)=Ae and ⌃(1 z)=A⇤e , with A⇤(x, y)=A(1 x, y)and⇤⇤(x, y)= ⇤(1 x, y). The zeros on the critical line correspond to the particular solution ⇤ = ⇤⇤ and 1 lim 0+ cos ⇤ =0.Thuslim 0+ ⇤ = n + ⌅ and replacing n n 2, the imaginary parts ⇥ ⇥ 2 ⇤ of these zeros must satisfy the exact equation⇥

1 i 1 3 arg + yn yn log ⇧⌅ +limarg ⇥ + iyn = n ⌅. (18) 4 2 0+ 2 2 ⇥ ⇥ ⇥ ⇥ The Riemann-Siegel ⌥ function is deﬁned by

⌥(t) arg 1 + i t t log ⇧⌅, (19) ⇥ 4 2 ⇥ where the argument is deﬁned such that this function is continuous and ⌥(0) = 0. Therefore, 1 there are inﬁnite zeros in the form ⇧n = 2 + iyn, where n =1, 2,..., whose imaginary parts

8 where i⇥(x,y) i⇥ (x,y) B(x, y) e + e 0 . (11) ⇤ + The second equality in (10) follows from A = A⇤. Then, in the limit 0 , a zero ⇧ corresponds to A =0,B =0orboth.Theycansimultaneouslybezerosincetheyare not independent. If B =0thenA =0,sinceA ⇥(z) . However, the converse is not ⌃ | | necessarily true. Since there is more structure in B, let us consider B =0.Thegeneralsolutionofthis

⇤ = ⇤⇤, lim cos ⇤ =0. (12) 0+ The converse of Riemann’s⇥ result On the critical line, the first equation (12) is already satisfied. Now, in the limit 0+, ⇧ the second equation implies ⇤ = n + 1 ⌅, for n =0, 1, 2,..., hence Recall Riemann’s main result: to calculate2 primes,± ± one y y ⇥ 5 1 needs to know the zerosn = of logzeta. +lim arg ⇥ 1 + iy + . (13) 2⌅ 2⌅e 8 0+ ⌅ 2 ⇥ ⇤ ⌅ ⇥ Our resultsA closer give inspection the converse: shows that the to right calculate hand side ofzeros, (13) has you a minimum need in the interval The particular solution( 2, 1), = thus, n cos(is bounded)=0 from givesan below, infinitenumberofzerosonthecriticalline i.e. n 1. Establishing the convention that to know all the primes. ⌅ 1 zeros are labeled by positive integers, ⇧n = 2 + iyn where n =1, 2,..., we must replace n n 2 in (13). Therefore, the imaginary1 parts of these zeros are determined from the staircase function,⇧ The however n th zero it is ofis the discontinuous form ⇥ = + atiyn the ordinates of non-trivial zeros, where it Recall solution of the transcendental equation 2 jumps by the multiplicity of the zero. Since ⇧(T ) is smooth, these jumps come from S(T ). trans. eqn: yn yn 1 1 11 log +lim arg ⇥ + + iyn = n (n =1, 2,...). (14) 2⌅ 2⌅e 0+ ⌅ 2 8 Now, if the EPF is valid, then⇥ there are no zeros to the right of the critical line. Then ⇤ ⌅ ⇥ S(T ) definedIn by short, piecewise we have shown integration that, asymptotically, does not there encounter are an infinite any number zeros of zeros as one on the approaches the critical line whose ordinates can be determined by solving (14) for yn. critical lineConjecture in theNote piecewise that, 3: by comparingThe integration, validity with the of counting andEPF must functionfor Re beN( the(zT)), > the same1/2 left hand as side of (14) is smoothsamonotonicincreasingfunctionof out S(t) and this leadsy, and to the a leading unique term solution is a smooth function. Possible discontinuities can only come fromS(1 Targ)=⇥ 1 + limiy ,S and(T in)(41) fact, it has a jump discontinuity to the transcendental equations⇤ for2 0each+ n. ⇥ 1 by one whenever y corresponds to a zero. However,⇥ if lim 0+ arg ⇥ + + iy is well ⇥ 2 where defined, then the left hand side of equation (14) is well defined for any y and there⇥ is a N unique solution1 for every1 n. Under this assumption,1 the number of solutions1 of/2 equation it By EPF: S(t) arg ⇥ + + it = lim log 1 pn . (42) ⇥ ⇤ 2 ⇤ N ⇧ ⇥⇤ ⇤n=1 ⌅ ⇥ 7 ⇧ ⇥ This is an explicit formula for S(T ) expressed as a sum over primes, and for strictly not zero, S (T ) is continuous. As explained in [? ], S(t) defined by this limiting procedure is Every individual zero knows about all the primes! also well-defined at the ordinate of a zero on the critical line. The function S(T )knowsabouttheRiemannzerossinceitjumpsateachzero.Thus,

the expression (38) for N(T ), with S(T )replacedbyS(T )in(42),whichinvolvesasum over primes, is a relation between Riemann zeros and the primes that is completely the inverse of Riemann’s result for the prime number counting function ⇤(x)expressedasa sum over non-trivial zeros. For the latter, one needs to sum over all zeros to identify the primes. Our result is the inverse; to ﬁnd the zeros, one must sum over all primes. In this sense the distribution of non-trivial zeros on the critical line is explicitly determined by

the prime numbers. In Figure 3 we plot equation (38) with S(T ) S(T ), given by (42), 3 ⌅ with a ﬁnite (small) number of primes. The jumps correspond to the nontrivial zeros of ⇥(s). A stronger version of this is presented below; see the discussion following (44). If one replaces p n log n, then S (T )nolongerjumpsatthezeros.Thisindicatesthatthezeros n ⇤ themselves and their GUE statistics [??] arises from ﬂuctuations in the primes. It would be very interesting to understand the origin of the GUE statistics in this way. S(t)=O(1) using the smooth approximation.

B. A transcendental equation for the n-th zero

1 Let us characterize precisely the zeros on the upper-half of the critical line, ⌅n = 2 + itn

for n =1, 2, 3,.... In [??] a transcendental equation for each tn was proposed which depends only on n. A more lengthy discussion of this result can be found in our lectures [?

18 Conclusions

• According to our conjectures, the validity of the RH needs both the EPF and the functional equation. • These two work together: The validity of the EPF and existence of solutions to the transcendental equations are closely related. • Known counter-examples to RH have no EPF. • We extended to another infinite class of L- functions based on modular forms. Brings in reasonably recent (1975) results of Deligne. • A unified perspective on different of L-functions Te End