Orthogonal Polynomilas, Xi-function And Riemann Hypothesis

ORTHOGONAL POLYNOMIALS, MOMENT PROBLEM AND THE RIEMANN XI-FUNCTION ξ (1/ 2 + iz ) Jose Javier Garcia Moreta Graduate student of Physics at the UPV/EHU (University of Basque country) In Solid State Physics Address: Address: Practicantes Adan y Grijalba 2 5 G P.O 644 48920 Portugalete Vizcaya (Spain) Phone: (00) 34 685 77 16 53 E-mail: [email protected]

ABSTRACT: In this paper we study a set of orthogonal Polynomials certain given measure

ω ( x ) = dα ( x )

{ pn ( x )} with respect a

related to the Taylor series expansion of the Xi-function

ξ (1/ 2 + iz ) , this paper is based on a previous conjecture by Carlon and Gaston related to the ξ (1/ 2) fact that Riemann Hypothesis (with simple zeros) is equivalent to the limit

lim

n →∞

p2 n ( z ) ξ (1/ 2 + iz ) = for a certain set of orthogonal Polynomials, we study the p2 n (0) ξ (1/ 2) ∞

‘Hamburger moment problem’

{µn }

here the moments

a2 n ( 2n ) ! = µ2 n = ∫ dα ( x ) x 2 n for even ‘n’ and 0 for n odd a0 −∞

are related to the power series expansion of Xi-function

∞

ξ (1/ 2 + iz ) a = ∑ ( −1) n 2 n z 2 n a0 ξ (1/ 2) n =0

, we also give the integral representation for the generating

∞

function

∑ ( −1)

n

µ2 n z 2 n = f ( z )

, in terms of the Laplace transform of

n =0

through all the paper we will use the simplified notation •

ξ (1/ 2 + iz ) ξ (1/ 2)

,

∞

∞

{ pn ( x )} = { pn ( x )}n =0 , {an } = {an }n =0

Keywords: Keywords: Orthogonal Polynomials, moment problem, Riemann Xi-function.

ORTHOGONAL POLYNOMIALS A set of Polynomials

{ pn ( x )} are orthogonal with respect to a certain measure on L2 (a, b) if

b

∫ dα ( x ) p ( x ) p n

m

( x ) = hnδ m ,n ( δ m ,n Kronecker delta ) , in this paper we will investigate a certain set

a

of Polynomials

{ pn ( x )} on ( −∞, ∞ ) respect to a certain even measure ω ( x ) = dα ( x ) = ω( − x )

1

,

we have used mainly the notation of Szego’s [7] book in order to descrbie the properties of these orthogonal Polynomials. We will study a set of orthogonal Polynomials with respect to a certain measure with the following properties ∞

•

The measure

ω ( x ) = dα ( x ) is positive on the interval ( −∞, ∞ )

,

∫ dα ( x ) > 0 with −∞

d α ( x ) ≥ 0 ,also it satisfies the functional equation ω ( x ) = ω ( − x ) .

•

The orthogonal Polynomials (Szego page 29) according to the property above will be odd or even depending on the value of argument ‘n’ p2 n +1 ( x ) = − p2 n +1 ( − x ) and for the even case

p2 n ( x ) = p2 n ( − x ) this fact is related to the definition of the moment problem µ2 k +1 = 0 for odd ‘k’ and µ2 k > 0 for k even pn ( x ) are REAL and distincts on the interval ( −∞, ∞ )

•

All the roots of

•

Since the measure is even, the ‘Hamburger moment problem’ associated to this measure , will ∞

have only non-zero even moment

∫ d α ( x )x

2n

= µ2 n > 0 the odd moment vanish µ2 n +1 = 0

−∞

µ0 = 1 , since in our case this ‘Hamburger moment problem’ is solvable (the Taylor series for the Xi-function is unique) for the measure ω ( x ) = dα ( x ) , the Hankel , for n=0 we can take

Matrix related to this moment problem defines a positive definite quadratic form n

∑ µi + juiu j = i , j =0

∞

∫ dα ( x ) ( u

0

2

+ u1 x + u2 x 2 + .... + un x n ) ≥ 0 with ui ∈ C (1)

−∞

In the paper [1] Carlon and Gaston showed the Riemann Hypothesis equivalence in term of orthogonal Polynomials

lim

n →∞

ω ( x ) = dα ( x )

p2 n ( z ) ξ (1/ 2 + iz ) = ,However they do not specify what the measure p2 n (0) ξ (1/ 2)

should be in order this formula to be true, our starting point is the taylor series for the

s − s 1 2  Xi function defined in terms of the Riemann Zeta as ξ ( s ) = s ( s − 1)π Γ   ζ ( s ) 2 2 ∞

1 ξ ( s ) = ∑ a2 n  s −  2  n =0

2n

2n ∞ d ( x 3/ 2 Ψ '( x ) )  1 4  −1/ 4 a2 n = dx  ln( x )  x dx (2n )! ∫1 2 

This formula (taylor expansion) is valid for

1 1 , if we set s = + iz and divide all by ξ (1/ 2) the 2 2

∞ ξ (1/ 2 + iz ) a = 1 + ∑ 2 n ( −1) n z 2 n ξ (1/ 2) n =1 a0

taylor series becomes variable

s<

(2)

, the integrand inside (2) under a change of

x = e 2u can be considered as a ‘Hamburger moment problem’ we define ∞

a2 n ( 2n )! = µ2 n = ∫ dα ( x ) x 2 n for even ‘n’ a0 −∞

and

µ2 n+1 = 0

  u d Ψ(e 2u )  d  e du  − u / 2  1  ω ( x ) = dα ( x ) =  if x ≥ 0 e du  ξ (1/ 2)  ω( x) = ω(− x)

2

(3)

∞

Ψ( x ) = ∑ e− n π x n =1

2

(4)

From this definition we can see that this measure is even and positive , in this case the ‘Hamburger moment problem’ can be solved by using the Taylor expansion of ξ (1/ 2 + iz ) near z =0 , the odd moment are 0 as one would expect by symmetry arguments and the first moment is normalized to be µ0 = 1 , this relation between ‘Hamburger moment problem’ and orthogonal Polynomials is important in

pn ( x ) , for example in

order to avoid tedious calculations of the n-th order orthogonal polynomial

general in order to obtain a orthogonal basis of Polynomials on a certain interval with respect to a given measure

d α ( x ) one can use the set of non-orthogonal powers of x {1, x, x 2 , x 3 ,.......} and then

compute

pn ( x ) using the ‘Gram-Schmidt’ orthogonalization procedure n −1

x n | pk ( x )

k =0

pk ( x ) | pk ( x )

pn ( x ) = x − ∑ n

p0 ( x ) = 1

pk ( x )

n ≥1

(5)

∞

Here the scalar product is defined via the Stieltjes integral

f |g =

∫ dα ( x ) f ( x ) g ( x ) −∞

pn ( x ) make these orthogonal polynomials hard to compute for practical calculations. A faster method to compute pn ( x ) without using the recurrence relation (5) is using the

However this definition for

Determinant of a Hankell Matrix whose entries are precisely the moments (in our special case) ∞

a2 n ( 2n )! = µ2 n = ∫ dα ( x ) x 2 n , µ2 n+1 = 0 in this case we can compute the n-th orthogonal a0 −∞ Polynomial as follows

pn ( x ) =

1 Dn Dn −1

pn ( x ) = 0

µ2

µ2 ....

0 .....

µn −1 µn

µn +1

1 0 .... 1

For n=0 then

1 Det ( H n | x ) with: Dn Dn −1

x

x

2

.... µn .... µn +1 .... .... ... 0 .... x n

1 0

0

µ2

µ2

0

Dn = ....

....

.....

µn −1 µn µn +1 µn µn +1 µn + 2

D−1 = 1 , (see reference [7] ) , in our case µ2 n +1 = 0 and

.... µn .... µn +1 .... ... ....

.... 0

(6)

µ2 n

a2 n ( 2n ) ! = µ2 n this makes a0

the Determinant easier to compute without using ‘Gram-Schmidt’ Orthogonalization for every positive ‘n’ , the relation to the conjecture proposed by Carlon and Gaston [1] is the fact that as n → ∞ then Riemann Hypothesis is equivalent to lim

n →∞

p2 n ( z ) ξ (1/ 2 + iz ) = , since all the roots of the orthogonal p2 n (0) ξ (1/ 2)

polynomials are real and distinct , and the polynomials must be odd or even functions, this hypothesis seems to be plausible unless there are multiple non-trivial zeros of the Riemann Zeta function ζ ( ρ ) = 0 = ζ '( ρ ) in this case the conjecture could be false but Riemann Hypothesis still remain true. Ullman [8] has studied the functional relation between orthogonal polynomials

pn ( x ) and the

2 n −1

determinant of a certain Hankell Matrix , if we define s2 n ( x ) =

∑a z k

( k −1)

to be the n-th section of a

k =0

Taylor power series then the distance

| s2 n ( x ) − p2 n ( x ) |→ 0 as n → ∞ , Carlon and Gaston [1]

3

studied the convergence of the Hadamard product for the Xi-function

ξ (1/ 2 + iz ) ∞  z2  = ∏ 1 − 2   γ  ξ (1/ 2) j =0  j 

,

p2 n ( x ) ,which we know to be real (and distincts) to these zeros of 1 the Xi-function over the critical line s = + iz , in this paper following their idea we have obtained an 2 even measure ω ( x ) = dα ( x ) = ω ( − x ) , this ansatz comes from the definitions (2) (3) (4) , from the the idea is to associate the zeros of

definition of ‘Hamburger moment problem’ we cand find using the Borel resummation procedure the following relations (generating function for the moments µ2n in terms of the Xi-function ) ∞

∞

∞

a2 n ξ (1/ 2 + izx ) ( −1) n (2n )! z 2 n = ∫ e − x dx ξ (1/ 2) n =0 a0 0

f ( z) = ∑

and

f ( z) =

dα ( x )

∫ 1 − ( izx )

2

(7)

−∞

(Laplace transform )

(Stietjles transform )

The argument for the formula (7) were obtained by us in a previous paper regarding divergent series [4] although we can prove (7) only in formal sense, in fact we have for the Laguerre or Legendre Polynomials with measure and moment

{s

n

= n ! , ω1 ( x ) = e − x } (Laguerre) and

1   , s 2n+1 =0 , ω2 ( x ) = 1 (Legendre) , formula (7) can be stated as  s2 n = 2n + 1   ∞

1

e −1/ z  1  ∞ e − x dx E1   = ∑ n !( − z ) n = ∫ z 1 + ( zx )  z  n =0 −∞

tan −1 ( z ) ∞ ( −1) n 2 n dx =∑ z =∫ (8) 2 z 1 ( zx ) + n =0 2 n + 1 −1

(Laguerre)

(Legendre)

These results were already known , in the case of Laguerre Polynomials, the integral representation is just the Borel resummed series for the exponential integral , expression (8) can be used to extend the domain of the power series to every real or complex ‘z’ (unless we have poles) , formula (7) can be used in order to obtain some information about the moment µ2n as z → ∞ using ∞

∞ dα ( x ) ( −1) k µ2 k ∫ x2 + z2 ≈ ∑ z 2k +2 k =0 −∞

a2 n ( 2n ) ! = µ2 n and µ2 n+1 = 0 a0

As a final remark, we have constructed a set of orthogonal Polynomials

(9)

pn ( x ) which are orthogonal

∞

with respect to a certain known measure

∫ dα ( x ) p ( x ) p n

m

( x ) = δ m ,n , the idea is if we could define a

−∞

1 dn n ω ( x ) (Q ( x ) ) , where Q(x) is a function that does not n d nω ( x ) dx p ( z ) ξ (1/ 2 + iz ) = and Riemann Hypothesis with simple depend on ‘n’ , if the conjecture lim 2 n n →∞ p (0) ξ (1/ 2) 2n

‘Rodrigues formula’

pn ( x ) =

(

)

zeros is true , then using Cauchy’s theorem we can give the following integral equation representation for Q(x) ,

p2 n (0) ξ (1/ 2 + ix ) (2n )! dzQ 2 n ( z ) here {d n } are some real constants and ‘C’ is a ≈ 2π i C∫ ( z − x )2 n +1 ξ (1/ 2) d 2 n −1

closed curve including the point z = x. Although many of the results were stablished in Cardon and Gaston’s paper he main difference with their paper is that they did not specify any measure

ω ( x ) for the scalar product of function f | g

4

involved

in the ‘Gram-Schmidt’ procedure neither they did give any Hankel determinant representation for the n-th orthogonal Polynomial in terms of any known ‘moment problem’ , we have exploited the advantages of having a symmetryc measure ω ( x ) = ω ( − x ) and the Hankel representation of orthogonal polynomials to simplify the Numerical calculations , we also have compared our results with other results referring Laguerre or Legendre Polynomials , our idea or conjecture is the following : the orthogonal polynomials

{ pn ( x )} that Cardon and Gaston are looking for are related to the solvable moment problem ∞

∫ d α ( x )x

2n

= µ2 n =

−∞

a2 n (2n )! > 0 (n even) , µ2 n +1 = 0 (n odd) ω ( x ) = dα ( x ) = ω ( − x ) , one a0

we have solved the problem and obtained the measure, we use the Hankel representation of the Polynomials

pn ( x ) =

1 Det ( H n | x ) to simplify the calculations to check if the conjecture in Dn Dn −1

paper [1] is correct. o

ξ (1/ 2 + iz ) :

Relation between our given measure and the Fourier and Laplace integral for

Although it could seem that we have chosen our measure at random in order to get the set

{ pn ( x )} ,

which is based only on the Taylor series expansion for the Xi-function ,we can also give a valid argument based on the Fourier (cosine) integral formula for ∞

Ξ( z ) = 4 ∫ duΦ (u) cos(uz )

Ξ( z ) = ξ (1/ 2 + iz )

∞

with

Φ (u ) = ∑ ( 2n 4π 2 e9 u / 2 − 3n 2π e5u / 2 ) e − n π e (10) 2

4u

n =1

0

The last expression inside (10) includes the first and second derivative of the Theta function ∞

Ψ ( x ) = ∑ e − n π x evaluated at x = e 4u , to extend the defintion of Φ to negative values of ‘u’ we 2

n =1

Φ (u ) = Φ ( −u ) , we have the relation between our measure and the u function involved in the Fourier Cosine transform (10) definition of ξ (1/ 2 + iz ) as Φ   = Aω (u ) 2 a , A is a real-valued constant that does not depend on the values of 2 n or (2n )! . a0 can impose the functional equation

∞

We can reformulate our initial moment problem as

∫ dxΦ ( x ) x −∞

2n

∝

a2 n (2n )! in terms of Φ ( u ) , a0 22 n ∞

using the Taylor series expansion for the cosine function

cos(2 zu) = ∑ ( −1)n n =1

( zu )2 n 22 n and (2n )!

integration term by term, we recover the well-knwon Taylor expansion for Xi-function 2n ξ (1/ 2 + iz ) ∞ n a2 n z = ∑ ( −1) a0 ξ (1/ 2) n =0

. Suprisingly this kind of relation also holds for the Chebyshev

Polynomials (with a change of variable

x = sin(t ) ) and for the Legendre Polynomials

π  x2  1 cos( xt ) J 0 ( x ) = ∏  1 − 2  = ∫ dt λn  π 0 n  1 − t2

1  sin( x ) x2  1 = ∏  1 − 2 2  = ∫ dt cos( xt ) (11) x n π  2 −1 n 

J 0 (λn ) = 0 n = 0,1, 2,3,..... Real roots of zeroth-order Bessel function, as it can be seen the functions represented in (11) by a Fourier cosine transform with a certain measure , describe real function with

5

ONLY real roots. This

Φ (u ) could be viewed as certain Probability distribution dF

x

2 4u  ∞ du ( 2n 4π 2e9u / 2 − 3n 2π e5u / 2 ) e−n π e  = F ( x) = Pr ob[Y ≤ x] (12) ∫−∞  ∑ n =1 

or if we define a G function via

G− ( z − ix ) + G+ ( z + ix ) = cos( zx ) and use the convolution definition

∞

( G * dF ) ( z ) = ∫ G( z − ix )dF ( x ) , then the Xi-function is proportional to the linear combination −∞

( G+ * dF ) ( z ) + ( G− * dF ) ( z ) , with

G± ( z ± ix ) defined on the upper/lower part of the complex

plane . Also In formula (7) we gave the expression for the moment generating function as the Laplace transform of the Xi function , now we extend by analytic continuation to define Laplace transform for complex values of ‘z’ via

z → ± si and use the definition of inverse Fourier transform , assuming f (i / s ) − f ( − i / s ) = Φ ( s ) valid for s ∈ R + , 8iπ s

has only real zeros , then we can write ∞

f ( z ) = ∑ ( −1)n (2n )! n =0

ξ (1/ 2 + iz ) ξ (1/ 2)

a2 n z 2 n . An alternative method to invert the Laplace transform a0

∞

f (1/ z ) ξ (1/ 2 + ix ) = ∫ dxe − zx Szego [7] is to define a set of secondary orthogonal polynomials via z ξ (1/ 2) 0 ∞

∫

the integral

−∞

p n ( t ) − pn ( x ) ω (t )dt = Qn ( x ) , then the quotient (Pade approximant) t−x

f ( i / z ) ∞ dα ( x ) Qn ( z ) fn ( z) = will converge to = ∫ 2 , then if we expand into power series z2 z − x2 pn ( z ) −∞ −s

Q2 n (is ) ∞ (2i )! = ∑ ( −1)i c2in 2 i +1 and perform a Laplace inverse transform to get the series p2 n (is ) i =0 s

∞

∑ ( −1) c i

2 in

x 2 i , the conjecture by Cardon and Gaston is equivalent to lim c2 in = n →∞

i =0 ∞

series

∑ ( −1)i i =0

a2 i 1 and a0 z 2i

∞

∑ (−1) n =0

i

µ2 i (2i )!

so both

c2 ni must converge (either in the usual sense for the sum or in z 2i

1 i ξ  + 

∞

 2 z  = z dxB( x )e − zx , (12) the sense of the Borel resummation for the divergent series ) to ∫0 ξ (1/ 2) ∞

B( x ) = ∑ i =0

( −1)2 i a2i 1 in the limit n → ∞ , this method of continued fractions and Pade (2i )! a0 x 2i

approximants applied to the moment problem is explained in [9] , expression (12) can be used to define the values of

Ξ( z ) = ξ (1/ 2 + iz ) for Real ‘z’ .

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CONCLUSIONS AND FINAL REMARKS

Inspired by a previous idea based on a formulation for Riemann Hypothesis as the limit

lim

n →∞

p2 n ( z ) ξ (1/ 2 + iz ) = and using the coefficients for the Taylor series of Xi-function p2 n (0) ξ (1/ 2)

ξ (1/ 2 + iz ) ∞ a = ∑ ( −1) n 2 n z 2 n ξ (1/ 2) a0 n =0

we have defined a ‘Hamburger moment problem’ based on the

u 2u ∞ 2 1 d ( e ∂ u Ψ( e ) ) − u / 2 e = ω ( x ) , Ψ ( x ) = ∑ e − n π x for x ≥ 0 , this definition measure du ξ (1/ 2) n =1 can be extended to negative values by using the functional equation ω ( x ) = ω ( − x ) ,with the solution to this moment problem we construct a set of orthogonal Polynomials { pn ( x ) } ,for even big ‘n’ the 1 Det ( H 2 n | x ) should approach to the Riemann XiHankel determinant p2 n ( x ) = D2 n D2 n −1

function divided by

ξ (1/ 2) = a0

as Cardon and Gaston deduced , from the measure ∞

the (asymptotic) series representation for

lim

n →∞

ω( x)

we can give

k

∞ dα ( x ) ( −1) µ 1 1 ≈ ∑ ∫0 x 2 + z 2 k =0 2 z 2 k +22k = 2 z 2 f  z  , in case

p2 n ( z ) ξ (1/ 2 + iz ) = holds , then Riemann Hypothesis would follow from the fact that the the p2 n (0) ξ (1/ 2)

z2  and the determinant Det ( H 2 n | x ) would have the same set of (real) 2   j =0  j  1  roots γ j ∈ R , so p2 n (γ ) = 0 = ξ  + iγ  γ ∈ R , Numerical computations can be simplified by the 2   fact that for a fixed ‘2n’ we can use the determinants of Hankel matrices to compute p2 n ( x ) ∞

infinite product



∏  1 − γ

References: [1] Cardon David and Gaston Sharleen “An equivalence for the Riemann Hypothesisin terms of Orthogonal Polynomials “Journal of approximation theory, ISSN 0021-9045, Vol. 138, Nº 1, 2006 [2] Bender Carl and Ben-Naim E. “Nonlinear-integral-equation Construction of Orthogonal Polynomials” Journal of Nonlinear Mathematical Physics , Vol 15 (2008) 73-80. [3] Conrey, J. B. "The Riemann Hypothesis." Not. Amer. Math. Soc. 50, 341-353, 2003 [4] Garcia J.J ; “A comment on mathematical methods to deal with divergent series and integrals” e-print avaliable at http://www.wbabin.net/science/moreta10.pdf [5] Keiper, J. B. "Power Series Expansions of Riemann's Xi-Function." Math. Comput. 58, 1992. [6] Shohat, J. A.; Tamarkin, J. D. (1943), “The Problem of Moments”, New York: American mathematical society, ISBN 0821815016.

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[7] Szego Gabor “ Orthogonal Polynomials” American Mathematical Society ISBN-10: 0821810235 [8] Ullman J. “Hankel determinants whose elements are sections of a Taylor series. Part I ” Duke Math. J. Volume 18, Number 3 (1951), 751-756. [9] Wall H.S (1948). “Analytic Theory of Continued Fractions” . D. Van Nostrand Company Inc.

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