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Integral Transforms and Special Functions, 25 Vol. 26, No. 2, 9 95, http://dx.doi.org/.8/652469.24.97393 Open problems Christian Berg Downloaded by [Copenhagen University Library] at :7 4 November 24 Department of Mathematical Sciences, University of Copenhagen, Copenhagen, Denmark The following open problems were presented at OPSFA-2, Sousse, Tunisia, March 29, 23. Christian Berg, University of Copenhagen, has collected these open problems. Mohamed J. Atia: Conjecture about a symmetric system of orthogonal polynomials We consider the weight w(x) = x + 2 / + x + x 2 / x on [, ], and claim that there exits a sequence of polynomials {P n } n orthogonal with respect to w(x) on [, ] and which fulfils P (x) =, P (x) = x () P n+2 (x) = xp n+ (x) γ n+ P n (x), n such that γ = 2, γ 2 = 4, γ 3 = 7 3, γ 4 = 4 5, that is to say γ 3 + γ 4 = 2, γ 5 = 4, γ 6 = 2 49, γ 7 = 25 98, γ 6 + γ 7 = 2, γ 8 = 4, γ 9 = 387 287, γ = 624 6435, γ 9 + γ = 2, γ = 2, γ 3n+2 = 4, γ 3n+3 + γ 3n+4 = 2. This conjecture was presented, first, in Leuven, Belgium 29, second, in Madrid, Spain 2 and, third, in Sousse, Tunisia 23. The weight w(x) is the solution of a second-order differential equation that is why we call the sequence {P n } n a sequence of second category. Email: berg@math.ku.dk c 24 Taylor & Francis
Integral Transforms and Special Functions 9 References [] Atia MJ, Leffet S. On orthogonal polynomials: semiclassical and of second category. Integral Transforms Spec Funct. doi:.8/652469.2.55362 University of Gabès, Tunisia jalel.atia@gmail.com Downloaded by [Copenhagen University Library] at :7 4 November 24 Howard Cohl: Hypergeometric series Given a hypergeometric series ( ) a,..., a p p+f p ; x, b,..., b p where the parameters in the hypergeometric series satisfy the relations a + = a + b = =a p + b p, then the series is called well-poised. If one has a well-poised hypergeometric series, where there exists some i p, so that a i = b i + = a 2 +, with a C \{, 2, 4,...}, then the hypergeometric series is called very-well-poised. I have a terminating very-well-poised 9 F 8 () hypergeometric series, a, a, a, n, n, 9F 8 2,2a + n,2a + n, a 2 +,, + n + a,+ n + a, a + 2, n a, n a, a ;. 2 Does anybody have any ideas on how I can sum this in closed form? National Institute of Standards and Technology, USA hcohl@nist.gov Anbu Swaminathan: Pick functions The function which is analytic in the slit domain C\[, ) and whose Taylor coefficients are the Hausdorff moment sequences also belong to the set of Pick functions [] which are related to moment problems. I am interested in the following problem. Problem To characterize the members of this class of Pick functions. The generalized polylogarithm [3] is defined by the normalized function p,q (a, b; z) = k= (( + a)p ( + b) q /(k + a) p (k + b) q )z k, z <, where, k + a, k + b, p, q are complex numbers such that Re p >, Re q >. Note that for a = b =, p + q = r C, Re r >, p,q (a, b; z) = L r (z), is the polylogarithm or Jonquière s function that further reduces to Riemann zeta function for z =. Numerical evidences suggest the following conjecture.
92 C. Berg Conjecture For p + q p 2 + q 2, the ratio p,q (a, b ; z) p2,q 2 (a, b ; z) belongs to the class of Pick functions. Downloaded by [Copenhagen University Library] at :7 4 November 24 Note that for p + q = a and p 2 + q 2 = b, a b, with a = b = a 2 = b 2 =, this result that L a (z)/l b (z) is in the class of Pick functions, is obtained in [4] using duality techniques of function theory. It would be interesting to see if the ratio given in the conjecture has the g-fraction as there exist a relation between the completely monotone sequences and the g-fractions, see [6]. Hence, it is important to identify the corresponding g-sequence for this ratio. Particular cases of this conjecture are proved in [5]. Further generalized polylogarithms are particular cases of the well-known generalized hypergeometric function [3] and since certain ratios of Gaussian hypergeometric functions [2] and their q-analgoues [5] can be written in terms of g-fractions, I propose Problem 2 Characterize the ratios of generalized (basic) hypergeometric functions so that the corresponding g-fractions lead to the geometric properties of Pick functions. References [] Berg C, Pedersen HL. Pick functions related to the gamma function. Rocky Mountain J Math. 22;32(2):57 525. [2] Küstner R. On the order of starlikeness of the shifted Gauss hypergeometric function. J Math Anal Appl. 27;334(2):363 385. [3] Mondal SR, Swaminathan A. Geometric properties of generalized polylogarithms. Integral Transforms Spec Funct. 2;2(9):69 7. [4] Roth O, Ruscheweyh S, Salinas L. A note on generating functions for Hausdorff moment sequences. Proc Am Math Soc. 28;36(9):37 376. [5] Swaminathan A, Tsygvintsev A. Pick functions and chain sequences for hypergeometric type functions. Preprint. [6] Wall HS. Analytic theory of continued fractions. New York: D. Van Nostrand Company; 948. Indian Institute of Technology, Roorkee, India mathswami@gmail.com Walter Van Assche: Minkowski s question mark function In 94, Hermann Minkowski introduced an interesting function, which he called the question mark function and he denoted its values by?(x). This notation with a question mark is somewhat confusing, so instead I will denote the function by q and I will only consider it on the interval [, ]. If < x <, then we can write x as a regular continued fraction x =, a i N {}. a + a 2 + a 3 +...
Integral Transforms and Special Functions 93 Downloaded by [Copenhagen University Library] at :7 4 November 24 The Minkowski question mark function at x is then defined as ( ) k+ q(x) = 2. 2 a +a 2 + +a k k= If x is a rational number, then the continued fraction is terminating and q(x) is given by a finite sum (see Figure ). By setting q() = and q() = one can show that q : [, ] [, ] is a continuous and increasing function, so that q is a probability distribution function on [, ] which defines a probability measure on [, ]. This distribution function has the property that q (x) = almost everywhere on [, ] so that the corresponding measure is singular and continuous. See [7] for some information about this function and references. The definition via regular continued fractions is not convenient for integration. A more convenient way is to define it as a fixed point of an iterated function system with two rational functions. One has 2 q q(x) = ( x 2 q x ( ) x, x x 2, ), 2 x and one can show that the sequence of probability distribution functions (q n ) n N, with ( ) x 2 q n, x x 2, q n (x) = ( ) x 2 q n, x 2 x, and q any probability distribution on [, ], converges uniformly to Minkowski s question mark function. This allows us to compute integrals by a limit procedure Salem s problem f (x) dq(x) = lim n f (x) dq n (x). A first open problem is about the Fourier coefficients of Minkowski s question mark function: α n = e 2inπx dq(x). The Riemann Lebesgue lemma tells us that Fourier coefficients of an absolutely continuous measure on [, ] tend to zero. The Minkowski question mark function is singularly continuous, so one cannot use the Riemann Lebesgue lemma. Nevertheless, the support of q is the full interval [, ] and it was proved by Salem [5] that q is Hölder continuous of order α = log 2/(2log(( 5 + )/2)) =.722. Furthermore, Salem showed that n n α k = O(n α/2 ), k= so that α n converges to zero on the average and there is the possibility that α n. This is the problem posed by Raphaël Salem [5,]: Do the Fourier coefficients of the Minkowski question mark function converge to? If not, what are lim inf n α n and lim sup n α n?
94 C. Berg.8.6 Downloaded by [Copenhagen University Library] at :7 4 November 24.4.2 Figure..2.4.6.8 The Minkowski question mark function. Orthogonal polynomials Consider the orthonormal polynomials for the Minkowski question mark function: with recurrence relation x p n (x)p m (x) dq(x) = δ m,n xp n (x) = a n+ p n+ (x) + b n p n (x) + a n p n (x), n, with p = and p =.Thesymmetryofqaround the point 2 implies that b n = for all 2 n N. Rakhmanov s theorem [4] tells us that for an absolutely continuous measure μ on [, ] for which μ > almost everywhere, one has a n 4 as n. In our case q = almost everywhere, so one cannot use Rakhmanov s theorem to deduce the asymptotic behaviour of the recurrence coefficients. However, it is known (see, e.g., [2,3,6]) that there exist discrete measures and singular continuous measures (on [, ]) for which the recurrence coefficients have the behaviour b n 2 and a n as n, so that they are in the Nevai class (for the interval 4 [, ]). The open problem is whether the Minkowski question mark function is such a singular continuous function for which the recurrence coefficients are in the Nevai class (for the interval [, ]), i.e. Is lim n a n = 4? If not, then what are lim inf n a n and lim sup n a n? Acknowledgements I like to thank S. Yakubovich for bringing the Minkowski question mark function and Salem s problem to my attention.
Integral Transforms and Special Functions 95 References Downloaded by [Copenhagen University Library] at :7 4 November 24 [] Alkauskas G. The Minkowski?(x) function and Salem s problem. C R Math Acad Sci Paris. 22;35(3 4):37 4. [2] Lubinsky DS. Singularly continuous measures in Nevai s class M. Proc Am Math Soc. 99;:43 42. [3] Van Assche W, Magnus AP. Sieved orthogonal polynomials and discrete measures with jumps dense in an interval. Proc Am Math Soc. 989;6:63 73. [4] Rakhmanov EA. On the asymptotics of the ratio of orthogonal polynomials. Mat Sb. 977;3(45):237 252 (in Russian); translated in Math. USSR Sb. 977;32:99 23. Rakhmanov EA. On the asymptotics of the ratio of orthogonal polynomials, II. Mat. Sb. 982;8(6):4 7 (in Russian); translated in Math. USSR Sb. 983;47:5 7. [5] Salem R. On some singular monotonic functions which are strictly increasing. Trans Am Math Soc. 943;53:427 439. [6] Totik V. Orthogonal polynomials with ratio asymptotics. Proc Am Math Soc. 992;4:49 495. [7] http://en.wikipedia.org/wiki/minkowski s_question_mark_function (March 29, 23). University of Leuven, Belgium walter@wis.kuleuven.be