Rational appoximations to the zeta function Keith Ball June 29, 27 Abstact This aticle descibes a sequence of ational functions which conveges locally unifomly to ζ The numeatos (and denominatos) of these ational functions can be expessed as chaacteistic polynomials of matices that ae on the face of it vey simple As a consequence, the Riemann hypothesis can be estated as what looks like a athe conventional spectal poblem but which is elated to the one found by Connes in his analysis of the zeta function Howeve the point hee is that the ational appoximations look to be susceptible of quantitative estimation Intoduction The Riemann hypothesis [R] states that the Riemann zeta function has all its nontivial zeos on the citical line {s : Rs = /2} (the tivial ones being at 2, 4, 6 and so on) The eal content of the conjectue is that ζ has no zeoes to the ight of the citical line The customay fomulation is equivalent by vitue of the functional equation poved by Riemann In ode to show that a holomophic function has no zeoes in (say) a half-plane it suffices to expess the function as a locally unifom limit of holomophic functions with no zeoes thee This aticle descibes a sequence of ational functions which begins (s ), s + 2(s ), 4s 2 + s + 9 6(s + 3)(s ), (s + 2)(3s 2 + s + ) 4(s 2 + 6s + )(s ), (s + 2)(72s 3 + 49s 2 + 93s + 25) 3(3s 3 + 29s 2 + 6s + 5)(s ), ()
We define the sequence as follows Fo each intege m we define and the coefficients (a m,j ) by ( p m (t) = ( t) t ) ( t ) 2 m p m (t) = ( ) j a m,j t j These coefficients ae escaled Stiling numbes of the fist kind but it is moe convenient fo us to use the diffeent indexation and scaling hee We then set F m (s) = a m,j B j s + j whee the B j ae the usual Benoulli numbes given by the geneating function and z e z = B j j! zj G m (s) = ( ) j a m,j j= s + j The ational functions in question ae the atios Fo example and F m (s) (s )G m (s) F 3 (s) = s 2s + 6(s + ) = 3s2 + s + 2(s )s(s + ) G 3 (s) = s 6s + s + 6(s + 2) = s 2 + 6s + 3(s )s(s + )(s + 2) It is immediate that the m th atio intepolates ζ at the points,, 2,, m and has a simple pole with esidue at s = We shall show that the sequence conveges locally unifomly to ζ, at least to the ight of the line {s : Rs = }, (with the obvious convention at s = ) The numeatos (and denominatos) of these ational functions can be expessed as chaacteistic polynomials of matices that 2
ae on the face of it vey simple One way to state this is that the numeato of the m th function is the deteminant of 2 3 m 2 2 3 m 3 2 3 m + ( s) 4 3 2 m m+ m m 2 As a consequence, the Riemann hypothesis can be estated as what looks like a athe conventional spectal poblem: to show that the specta of cetain matices stay to the left of the citical line, at least asymptotically as m Needless to say, I spent some time thinking about this poblem without success but am cetain that I have not exhausted the possible lines of attack Even if the simplicity of these matices is indeed an illusion, as one would expect, thee ae concete easons to think that these ational appoximations to zeta might be useful fo estimating the size of zeta, in the sense of the Lindelöf hypothesis: see fo example the book of Patteson [P] Polyá suggested that the Riemann hypothesis should be poved by expessing the zeoes of zeta (otated onto the eal line) as eigenvalues of a self-adjoint opeato (The statement is often cedited to Hilbet: it was Tey Tao who pointed out the mistake to me) A numbe of candidates fo such opeatos have been poposed, coming mainly fom quantum theoy The best known of these was found by Connes [C] Thee is a connection between Connes infinite-dimensional opeato and the finite-dimensional ones descibed hee, which will be explained biefly in Section 5 This became appaent to me fom the vey eadable aticle of Lachaud [L] My hope is that the finite-dimensional opeatos and the ational functions they coespond to, ae susceptible of quantitative estimation that would not make sense fo the infinite-dimensional opeato It is well known that the zeoes of ζ should be modelled by the eigenvalues of cetain andom matices This oiginated in the wok of Dyson and Montgomey [M] and was expeimentally confimed by emakable calculations of Odlyzko [O] Katz and Sanak extended the model to othe L-functions [KaSa] In the past two decades a huge amount of wok has been done on this connection in paticula by 3
Keating, Snaith and thei collaboatos [KeSn] While stictly speaking this is only indiectly elated to the esults in this aticle the andom model is clealy a cucial inspiation In ode to pove the convegence of F m(s) (s )G m(s) F m (s) h s m Γ(s)ζ(s) to ζ(s) we shall show that and (s )G m (s) h s m Γ(s) whee h m is the patial sum m j= /j of the hamonic seies We can get a sense of why this is, quite easily Using the following vaiant of Konecke s fomula B j = ( ) j j k= it is easy to show that fo Rs > F m (s) = ( m k= k + k + k = k = If x is close to zeo then p m (x) is appoximately ( ) k ( ) ( + ) j (2) ( ) ) k ( ) p m (( + )x) x s 2 dx (3) m e x/i = e hmx i= So the sum ove in equation (3) is appoximately k =( ) ( k ) e hm(+)x = e hmx ( e hmx ) k Thus fo small values of x the integand in equation (3) is appoximately ( m k= ) k + e hmx ( e hmx ) k x s 2 If the appoximation wee good fo all x between and then F m (s) would be close to k= k + e hmx ( e hmx ) k x s 2 dx = hm s hm k= k + e y ( e y ) k y s 2 dy 4
The last integal plainly conveges to k= k + e y ( e y ) k y s 2 y dy = e y e y y s 2 dy as m povided Rs > and the latte is easily seen to be Γ(s)ζ(s) Ou fist aim will be to show that indeed h s m F m (s) Γ(s)ζ(s) locally unifomly fo Rs > (not just Rs > ) as m This looks like a tall ode Cossing the pole at s = is not the poblem The difficulty is that unless x is vey close to, the expessions m,k (x) = k = ( ) k ( ) p m (( + )x) involve values of p m at points well outside the inteval [, ], whee p m looks nothing like a negative exponential Indeed m,k (x) is a divided diffeence of p m and consequently equal to ( x) k p (k) m (u) fo some u between x and (k + )x that is not easily specified Since p m oscillates epeatedly on the inteval [, m + ] it would seem that m,k (x) could be vey lage in size and of moe o less andom sign So the following lemma comes as something of a shock Lemma If m is a non-negative intege and p m (x) = ( x)( x/2) ( x/m) then fo each intege k and each x [, ] ( ) k k m,k (x) = ( ) p m (( + )x) = It is tivial to check that m,k (x) = k= fo all x, so the lemma shows that fo each m the m,k fom a patition of unity on [, ] and thus automatically contols the sizes of the m,k as well as thei signs Once the lemma is established the convegence poof is faily staightfowad: this will be the content of Section below The obvious way to pove the Konecke fomula (2) mentioned above is to use the expansion y e y = k= 5 ( e y ) k k +
that aleady appeaed in the integal fomula fo Γ(s)ζ(s) So it might be logically moe easonable to define the F m by using the fomula ( m ) F m (s) = k + m,k(x) x s 2 dx k= and simply avoid mention of the Benoulli numbes Howeve it seemed a little odd to define a ational function with known poles and esidues as the analytic continuation of an integal The convegence poof just alluded to elies on the fact that the F m ae defined as sums which we can pass though integal signs The point of the second section of the aticle will be to povide a bidge between the definition of the F m and thei epesentation as chaacteistic polynomials: in othe wods to epesent the F m as something moe like a poduct than a sum The main fomula in Section 2 is the following ecuence fo the F m : fo each m Thus and so on (s + m )F m (s) = m m + + (m + ) F m j (s) j(j + ) (s )F (s) = sf (s) = 2 + F (s) j= (s + )F 2 (s) = 3 + 3 2 F (s) + 3 6 F (s) If we teat the fist m + of these elations as a linea system fo the values F (s), F (s),, F m (s) we can expess the fact that F m (s) = by the vanishing of a cetain deteminant In Section 3 shall show that this deteminant can be witten as det(l m + ( s)u m ) whee as mentioned ealie L m is the (m + ) (m + ) Toeplitz matix 2 3 2 4 3 2 m+ m m 2 6
and U m is the matix 2 3 m 2 3 m 3 m m If we set s = z/(z ) the deteminant becomes det(zl m (L m +U m )) (apat fom a facto of ( z) m+ ) The Riemann hypothesis would follow if this deteminant vanishes only at points with z o equivalently that the matix I m+ + L m U m has spectal adius at most, whee I m+ is the (m+) (m+) identity matix Fo small values of m this is tue In the fist vesion of this aticle I stated that thee ae good easons to believe that the zeoes of the F m do leak acoss the citical line (and then come back again) Pace Nielsen quickly confimed that when m = 643 thee is a zeo to the ight of the citical line He also infomed me that when m = 27 thee is a zeo with eal pat lage than (which I had not expected) The Riemann hypothesis is equivalent to the statement that the spectal adius of I m+ + L m U m is at most + o() as m In Section 4 of the aticle I shall explain why I think that the appoximations F m (s) might be useful to estimate the size of ζ The main point is that wheeas appoximations to zeta that ae sums of powes oscillate wildly all the way up the citical line, a polynomial of degee m cannot oscillate too often Wheneve one has a new sequence of appoximations to ζ it is natual to ask whethe they can help to pove Diophantine popeties (iationality o tanscendence) of values of the zeta function and most especially Eule s constant ( γ = lim ζ(s) ) s s Since the appoximations descibed hee ae ational functions (with intege coefficients) they do povide ational appoximations to Eule s constant but fo this paticula value of zeta the appoximations ae not new 7
Acknowledgements I am extemely gateful to David Peiss fo his advice duing this wok My thanks also to Tey Tao and Pete Sanak who made vey helpful suggestions concening the pesentation The key lemma and convegence In the intoduction we defined, fo each m, p m (x) = ( x)( x/2) ( x/m), and fo each k m,k (x) = k = ( ) k ( ) p m (( + )x) Note that the sum makes sense and is zeo if k < o k > m We also intoduced the function F m (s) as a ational function, F m (s) = We also mentioned the Konecke fomula B j = ( ) j j k= k + a m,j B j s + j k = ( ) k ( ) ( + ) j It is a consequence of standad popeties of the binomial coefficients that fo all j between and m, the sum on the ight is unchanged if the uppe limit is inceased fom j to m This implies that F m (s) = = Fo Rs > the sum ove j can be witten as ( ) ( ) j a m,j k k ( ) ( + ) j j= s + j k= k + = ( ) k k ( ) m ( ) j a mj( + ) j k= k + = j= s + j ( ) j a mj ( + ) j x s+j 2 dx = p m (( + )x)x s 2 dx j= and so fo Rs > we have F m (s) = k= m,k (x) k + xs 2 dx 8
Fom now on we use f m (x) to denote m k= m,k (x) k+ The aim of this section is to pove the following theoem Theoem 2 hm s (s )F m (s) (s )Γ(s)ζ(s) locally unifomly fo Rs > (with the obvious convention at s = ) It is easy to check that fo Rs > Γ(s)ζ(s) = y e y e y y s 2 dy If we set u = e y fo y >, then since < u <, Theefoe y log( u) = e y u Γ(s)ζ(s) = k= = k= u k k + = k= ( e y ) k k + k + e y ( e y ) k y s 2 dy Most of the effot in poving Theoem 2 will go into showing that the tuncated functions f m (x/h m ) [,hm] convege to x e x = k + e x ( e x ) k k= on [, ) with the convegence dominated by a negative exponential function It is clea that fo each fixed x, p m (x/h m ) e x and hence that fo each fixed k and x m,k (x/h m ) e x ( e x ) k (4) as m We need to establish two types of dominance: one to confim that the sum f m (x) = k= m,k (x) k + conveges pointwise in x to k= k+ e x ( e x ) k and one to check that this convegence is dominated on [, ) Fo almost evey estimate we make it is essential to have the key lemma stated in the intoduction: 9
Lemma If m is a non-negative intege and p m (x) = ( x)( x/2) ( x/m) then fo each intege k and each x [, ] m,k (x) = k = ( ) k ( ) p m (( + )x) We also need a simple popety of the divided diffeences that depends only upon the fact that p m is a polynomial of degee at most m Lemma 3 If j is a non-negative intege then fo evey x, (k + )(k + 2) (k + j) m,k (x) = j!p m ( jx) k= In paticula fo j = m,k (x) = p m () = k= Poof We shall confim that fo any polynomial q of degee at most m ) k (k + )(k + 2) (k + j) k= =( ) ( k q(( + )x) = j!q( jx) and in checking this we may assume that x = So ou aim is to veify that fo each such q ) k (k + )(k + 2) (k + j) k= =( ) ( k q( + ) = j!q( j) It suffices to check this fo each polynomial of the fom q n : t (t )(t 2) (t n) with n m The intenal sum vanishes if q has degee less than k and hence it vanishes fo q n if k > n It also vanishes if k < n because of the fom of q n The only emaining case is k = n and in that case the intenal sum has value ( ) n n! So the double sum is (n + )(n + 2) (n + j)( ) n n! = ( ) n (n + j)! = j!q n ( j)
The poof of Lemma involves the intoduction of an additional paamete as follows Fo each v define and P m (v, x) = m!p m (x v) = (v + x)(v + 2 x) (v + m x) m,k (v, x) = k = ( ) k ( ) P m (v, ( + )x) Obseve that P m (, x) = m!p m (x) and m,k (, x) = m! m,k (x) So Lemma follows fom: Lemma 4 If m is a non-negative intege, k is an intege, v and x m,k (v, x) Poof We use induction on m When m =, m,k (v, x) is zeo unless k = in which case it is We claim that fo m > m,k (v, x) = (v + x) m,k (v +, x) + kx m,k (v + x, x) Once this is established the inductive step is clea because we can assume that k and fo the given ange of v and x, the numbe v + x is also at least Now fo any v and x P m (v, x) = (v + x)p m (v +, x) and so m,k (v, x) = k = ( ) k ( ) (v + ( + )x)p m (v +, ( + )x) ( ) k k = ( ) (v + x)p m (v +, ( + )x) = ( ) k k ( ) xp m (v +, ( + )x) = ( ) = (v + x)) k k m,k (v +, x) k ( ) xp m (v +, ( + )x) = (v + x)) m,k (v +, x) + kx = k = ( ) k ( ) P m (v +, ( + 2)x) = (v + x) m,k (v +, x) + kx m,k (v + x, x)
whee the last step follows fom the fact that fo all m, v and x, P m (v +, ( + 2)x) = P m (v + x, ( + )x) By combining Lemmas and 3 we can immediately make some estimates fo the m,k (x) that will give us pat of the dominance we need to get convegence Lemma 5 Fo each m, each k and each x [, h m ] and fo each m m,k (x/h m ) f m (x/h m ) p m (x/h m ) = k= (k + ) ex k + m,k(x/h m ) e x Poof Fo the fist one we apply Lemma 3 with j = and use the positivity of the m,k to deduce that fo each k m,k (x) k + p m( x) = k + ( + x)( + x/2) ( + x/m) (k + ) ehmx Fo the second one we obseve that k= k + m,k(x) k m,k (x) = p m ( x) p m () e hmx k= As aleady emaked it is clea that fo each fixed x, p m (x/h m ) e x and hence that fo each fixed k and x as m Fom Lemma 5 we have m,k (x/h m ) e x ( e x ) k (5) m,k (x/h m ) (k + ) ex 2
so the convegence in (5) is dominated (on the space of non-negative integes with counting measue) by a sequence summable against (/(k + )) Hence fo each x > f m (x/h m ) = k= k + m,k(x/h m ) x e x We have that hm s hm F m (s) = f m (x/h m )x s 2 dx In ode to use dominated convegence on [, ) we need an estimate fo f m which we pove by intoducing anothe exta paamete Fo each m and p define f m (p, x) = and obseve that f m (x) = f m (, x) Lemma 6 Fo each m and p k= k + + p m,k(x) f m (p, x) = f m (p, x) + px m f m (p, x) (p + )x m f m (p +, x) Poof As long as p > we have f m (p, x) = = k= m,k (x) ( + u) K m,x (u) ( + u) p+ du k+2+p du whee K m,x (u) = k= ( + u) k+ m,k(x) This function is holomophic on the plane apat fom its pole at Its deivatives at ae successively m,k (x) k= (k + ) m,k (x) k= (k + )(k + 2) m,k (x) k= 3
and so on and theefoe by Lemma 3 its powe seies expansion at is Theefoe, fo u < ( ) j p m ( jx)u j j= K m,x (u) = ( ) j p m ( jx)u j j= = ( ( ) j p m ( jx) + jx ) u j j= m = K m,x (u) + x ( ) j p m ( jx) ju j m j= = K m,x (u) + x m uk m,x(u) The outemost identity continues analytically so it holds fo all u > Theefoe f m (p, x) = K m,x (u) ( + u) = f m (p, x) + x m = f m (p, x) x m = f m (p, x) + x m du p+ uk m,x(u) ( + u) K m,x (u) d u du ( + u) ( K m,x (u) p+ du p+ du p ( + u) p + p+ ( + u) p+2 = f m (p, x) + p m x f m (p, x) p + m x f m (p +, x) ) du Now we can estimate f m (x) as follows Using the key lemma we have an inequality f m (, x) = k= k + 2 m,k(x) 2 k= k + m,k(x) = 2 f m(x) povided x Then by Lemma (6) f m (x) = f m (, x) = f m (, x) x m f ( m (, x) f m (x) x ) e x/(2m) f m (x) 2m So by induction f m (x) e hmx/2 4
and we get the negative exponential dominance f m (x/h m ) e x/2 on the ange of integation [, h m ] This suffices to guaantee that fo Rs > hm s hm F m (s) = f m (x/h m )x s 2 dx Γ(s)ζ(s) We wish to coss the pole and so we need to modify the integand Fo Rs > we have ( Γ(s) ζ(s) ) s = = x e x xs 2 dx ( x e x e x ) x s 2 dx e x x s 2 dx The last integand behaves like x nea so the integal conveges locally unifomly fo Rs > and epesents the holomophic function Γ(s) ( ) ζ(s) s on this lage egion As in the intoduction we set and obseve that fo Rs > G m (s) = ( ) j a m,j j= s + j So on this half-plane G m (s) = p m (x)x s 2 dx h s hm m G m (s) = p m (x/h m )x s 2 dx Now p m (x/h m ) e x fo each fixed x and p m (x/h m ) e x as long as x h m so fo Rs > hm p m (x/h m )x s 2 dx Γ(s) s Theefoe, still only fo Rs >, we have h s hm ( m (F m (s) G m (s)) = (f m (x/h m ) p m (x/h m )) x s 2 dx Γ(s) ζ(s) ) s The integand is dominated as x by e x/2 but also, as x by e x owing to Lemma 5 Moeove, F m and G m both have esidue at s = so the diffeence 5
is holomophic fo Rs > So the integal epesents h s m (F m (s) G m (s)) on the lage egion and also conveges to Γ(s) ( ) ζ(s) s on this egion To complete the poof of Theoem 2 it suffices to show that (s )h s m G m (s) Γ(s) locally unifomly fo Rs > We have that fo Rs >, (s )h s hm m G m (s) = (s ) p m (x/h m )x s 2 hm dx = /h m p m(x/h m )x s dx because p m () = The latte integal conveges as long as Rs > so it epesents (s )hm s G m (s) on the lage egion It suffices to show that /h m p m(x/h m ) [,hm] e x fo x > and that the convegence is dominated by a negative exponential Obseve that p m is deceasing on [, ] so p m(x) is positive fo x < On the othe hand p m m(x) = p m (x) j= j x ( x/j) + p m (x) j=2 j=2 j e hmx ( + h m ) Thus fo x h m giving the equied dominance Also /h m p m(x/h m ) = h m /h m p m(x/h m ) e x (/h m + ) 2e x m ( x/(jh m )) + p m (x/h m ) j=2 h m j=2 j x/h m The fist tem is at most h m e (hm )x/hm and tends to while the second tem behaves like e x (h m )/h m and tends to e x This establishes Theoem 2 Finally we have that the atios F m (s) (s )G m (s) convege locally unifomly to ζ(s) fo Rs > My guess is that they do so on the entie complex plane 6
2 The bidge fom sum to deteminant The pupose of this section is to establish the ecuence (s + m )F m (s) = m m + + (m + ) F m j (s) j(j + ) fo m which will enable us to expess the numeato of F m as a deteminant A simila ecuence holds fo the functions G m : if m (s + m )G m (s) = (m + ) j= j= G m j (s) j(j + ) A small modification of the poof below actually yields this as well We begin with a simple emak Lemma 7 Fo each non-negative intege m we have lim sf m(s) = s m + Poof lim sf m(s) = f m () s All the m,k vanish at x = apat fom m,m since they involve only values of p m at the integes, 2,, m By Lemma 3 the m,k add up to so m,m () = So f m () = k= m,k () k + = m + Now fo the main lemma of the section Lemma 8 Fo each non-negative intege m (s + m )F m (s) = m m + + (m + ) F m j (s) j(j + ) j= Poof The two sides have the same limits at infinity so it suffices to check that they have the same esidues at each of the points,,,, 2 m At s = the esidue on the left is (m )a m, B while the esidue on the ight is (m + ) j= 7 a m j, B j(j + )
It thus suffices to check that fo each m, (m )a m, = (m + ) j= a m j, j(j + ) (The case = m is also obvious) Multiplying by x and summing ove m it is enough to check that m p m (x) xp m(x) = (m + ) j= p m j (x) j(j + ) Both sides ae polynomials of degee m so we need only check the values at x =, 2,, m Let k be one of these integes p m (k) = and it is easy to check that On the othe hand (m + ) j= p m j (k) j(j + ) k p m(k) = ( ) k ( m k ) = (m + ) m j=m k+ j(j + ) m j because p m j vanishes at k if m j k The latte expession is (m + ) j=m k+ ( ) m j j(j + ) = (m + ) = (m + ) ( ) k m j k = = ( ) k (m + ) i= ( k i k ( ) ( ) k = (m + ) = (m )(m + ) ( ) k ( ) (x m x m ) dx ( x m x) k dx x m k ( x) k dx and the beta integal gives the appopiate ecipocal of the binomial coefficient The ecuence elation given by Lemma 8 descibes a dynamical system fo the sequence F m (s) Numeically this system appeas to evolve vey slowly and indeed the convegence of the sequence is vey slow This makes the appoximations useless fo effective calculation of the value ζ(s) but suggests that it might be possible to tack the dynamical system: it is almost a continuous-time system 8 )
3 The spectum Fom the pevious section we have that fo each m (s + m )F m (s) = m m + + (m + ) F m j (s) j(j + ) (6) j= The fist m + of these elations give us the linea system s 2 2 s 3 3 s + 6 2 m+ m(m+) m+ 6 m+ 2 s + m F (s) F (s) F m (s) = 2 m+ So F m (s) can be witten as the atio of two deteminants The denominato is the deteminant of the (m+) (m+) matix on the left, namely (s )s(s+) (s+ m ) The numeato is the deteminant of the matix obtained by eplacing the last column of the oiginal matix with the vecto (, /2,, /(m + )) It will be moe convenient to move this vecto to the fist column of the matix thus intoducing a facto of ( ) m into the deteminant, and to change the sign of all the othe columns, thus emoving the facto again Then the numeato can be witten det s 2 3 m m+ 2 s 2 3 3 s 6 2 m m (m )m 6 m+ m+ m(m+) 6 m s m + 2 2 Regad the columns as labelled,,, m We leave the zeo and columns unchanged as (, /2, /3,, /(m + )) and ( s,, /2,, /m) espectively We add the column to the 2 column to get (( s), ( s), 2,, 2/3,, 2/(m )) m+ 2 9
We add the (new) 2 column to the 3 column and we get (( s), ( s), ( s), 3, 3/2, 3/3,, 3/(m 2)) Continue in this way and afte all the additions divide the 2 column by 2, the 3 column by 3 and so on to get F m (s) = m! det(l m + ( s)u m ) (s )s (s + m ) (7) whee and L m = 2 3 2 4 3 2 m m 2 m+ 2 3 m 2 3 m 3 m U m = (9) m Notice that F m does not have poles at 2, 4, because the coesponding Benoulli numbes vanish So the deteminant in the numeato picks up the tivial zeos of zeta at these numbes: indeed the facto s + 2 appeas as soon as m 3, the facto s + 4 as soon as m 5 and so on The zeoes of F m ae thus elated in a simple way to the spectum of L m U m This fomulation fo F m is pehaps the most elegant one in tems of a deteminant but it is inteesting to expess F m in a fom in which the poblem looks moe like a conventional spectal poblem To begin with we multiply column j by j fo each j but leave the zeo column unchanged This includes the m! facto appeaing in (7) into the deteminant We now subtact the m ow fom the last, the m 2 ow fom the m and so on (8) 2
to poduce the matix 2 6 2 2 3 2 6 2 m(m+) (m )m 2 m (m 2)(m ) 2 +( s) We now add all the ows below the top one, to the top one, so that the second matix now has a zeo top ow The fist matix now has top ow ( m +, m, 2 m,, m ) Since the vaiable s now appeas on the diagonal in all places except the fist, ou aim is to educe the dimension by one so as to ceate a chaacteistic polynomial pope We add multiples (m + )/2, (m + )/6 and so on of the top ow to the successive ows below Since this eliminates the fist column below the fist ow, the deteminant is now the top left enty /(m + ) multiplied by the deteminant of the emaining squae So we get whee T m and R m ae as follows T m = det(t m + R m + (s )I m ) m + 2 2 6 2 2 3 (m )m 2 m m (m 2)(m ) 2 which is lowe tiangula with enty j in the jj diagonal place and enty j/((i j)(i j + )) in the ij place, if i > j R m is the ank one matix given by ( ) (m + )j R m = i(i + )(m j + ) 2 ij
We have F m (s) = m + det(t m + R m + (s )I m ) mj= (s + j ) We ae thus inteested in the spectum of the matix T m + R m whee T m is a cetain lowe tiangula matix and R m has ank It will be seen that this fomulation is actually somewhat close to the ecuence elation (6) If we set A m = T m + R m and B m = T m + R m I m then the deteminant in question is det(sa m (s )B m ) The complex numbes to the ight of the citical line ae those fo which s > s so a natual way to tackle the spectal poblem would be to ty to find a nom on C m with the popety that fo evey z C m A m z B m z The obvious choice would be an Hilbetian nom So we look fo a positive definite matix H fo which A mha m BmHB m is also positive definite o altenatively one fo which A m HA m B m HBm is positive definite The fom of the matix T m makes this vey tempting If we ignoe the ank one matix R m then we can cetainly compute the spectum of T m since it is lowe tiangula Howeve thee is a natual choice of nom which shows that the spectum is to the left of the citical line and theefoe povides a moe obust agument that one could ty to petub If H is the diagonal matix with enties, /2, /3,, /m on the diagonal then T m HTm (T m I m )H(T m I m ) = T m H + HTm H 22
is the matix 2 6 2 6 2 (m )m (m )m 3 2 2 (m 2)(m ) 5 3 2 (m 2)(m ) 2 2 m This is obviously positive definite because it has negative off diagonal enties and ow sums that ae positive because of the familia telescoping sum n= n(n + ) = m + As emaked ealie thee ae good easons to think that the zeoes of the F m do leak though the citical line so that the best one can hope fo is to find matices H m with positive definite and ɛ m ( + ɛ m )A mh m A m B mh m B m Once one is in possession of the matices T m and R m one could confim that they yield the F m in a diect way Diagonalise T m as P DP and check that when you apply P and P to the components of R m you ecove the Benoulli and Stiling numbes Such an agument would be a bit hash on the eade since thee would be little motivation fo intoducing these paticula matices Moe impotantly the dynamical system descibed by the ecuence elation (6) is of inteest in itself 4 Estimating the size of ζ Numeical evidence indicates that the function f m (x/h m ) diffes fom x/(e x ) by only about h m /m at any point of [, h m ] and so we expect the atio F m (s) (s )G m (s) to povide a good appoximation to ζ at s = /2 + it as long as Γ(s) is as lage as h m /m This happens if t is at most a bit less than 2 log m In fact, numeical π evidence and ough calculations indicate that the atio is not too fa fom ζ fo t all the way up to log m At the same time thee ae good easons to think that F m (s) 23
does not oscillate significantly fo t lage than log m So we have the tantalising possibility that the two egions ovelap: the t < log m egion whee F m tells us about ζ and the t > log m egion whee F m is smooth enough to be estimated This discussion suggests that one should look at the asymptotic expansion fo F m (s) which stats off F m (s) = (m + )(s ) + c m (m + )(s )s + whee the coefficient c m gows logaithmically with m, the next coefficient like (log m) 2 and so on Howeve my feeling is that the moe pomising appoach is the usual one: to look at an integal (say) F m (s) G m (s) = (f m (x) p m (x))x s 2 dx and move the contou into the egion whee x s 2 is vey small if s = /2 + it with t lage Fo the genuine ζ integal x e x e x x s 2 dx this appoach is hopeless because the contou is foced up against the imaginay axis and hence picks up the poles of z z e z Being a polynomial, z f m (z) has no poles so the issue does not aise The poblem is to estimate f m off the eal line 5 The connection with Connes opeato The Toeplitz matix L m given in equation (8) can be thought of as acting on polynomials a + a x + a 2 x 2 + + a m x m athe than sequences (a,, a m ) It does so by multiplication by the patial sum x j j + 24
of the seies fo log( x) (followed by tuncation back to a polynomial of degee m) x In this context the uppe tiangula matix U m in (9) maps the constant function to and fo each k the monomial x k to the sum + x + + x k k = xk k( x) Thus fo any polynomial q of degee m the image is q(t) q() dt x x t The opeato of Connes is built fom a multiplication opeato and an integal opeato much like these, acting on an infinite-dimensional function space Refeences [C] [KaSa] [KeSn] [L] [M] [O] [P] A Connes, Tace fomula in noncommutative geomety and the zeos of the Riemann zeta function, Sel Math, New Se 5 (999) N M Katz and P Sanak, Random matices, Fobenius eigenvalues, and monodomy, Colloquium publications, bf 45, Ameican Mathematical Society, Povidence, RI, (999) JP Keating and NC Snaith, Random matix theoy and ζ(/2 + it), Comm Math Phys, 24 (2) G Lachaud, Spectal analysis and the Riemann hypothesis, J Comp Appl Math, 6 (23) HL Montgomey, The pai coelation of the zeta function, Poc Symp Pue Math, 24 (973) A M Odlyzko, The 22 -nd zeo of the Riemann zeta function, Dynamical, Spectal, and Aithmetic Zeta Functions (M van Fankenhuysen and ML Lapidus, eds), Contempoay Math, Ame Math Soc, Povidence, RI, (2) S J Patteson, An intoduction to the theoy of the Riemann Zeta- Function, Cambidge studies in advanced mathematics 4, CUP (988) 25
[R] B Riemann, Übe die Anzahl de Pimzahlen unte eine gegebenen Gösse, Monatsb de Beline Akad, (858) 26