aa r X i v : . [ m a t h . N T ] J u l SECANT ZETA FUNCTIONS
MATILDE LAL´IN, FRANCIS RODRIGUE, AND MATHEW ROGERS
Abstract.
We study the series ψ s ( z ) := P ∞ n =1 sec( nπz ) n − s , and prove that it convergesunder mild restrictions on z and s . The function possesses a modular transformationproperty, which allows us to evaluate ψ s ( z ) explicitly at certain quadratic irrationalvalues of z . This supports our conjecture that π − k ψ k ( √ j ) ∈ Q whenever k and j arepositive integers with k even. We conclude with some speculations on Bernoulli numbers. Introduction
Let ζ ( s ) denote the Riemann zeta function. It is well known that ζ (2 n ) π − n ∈ Q for n ≥
1. Dirichlet L -functions and Clausen functions are modified versions of the Riemannzeta function, which also have nice properties at integer points [Lew91]. Berndt studieda third interesting modification of the Riemann zeta function, namely the cotangent zetafunction [Ber76]: ξ s ( z ) := ∞ X n =1 cot( πnz ) n s . (1.1)He proved that (1.1) converges under mild restrictions on z and s , and he produced manyexplicit formulas for ξ k ( z ), when z is a quadratic irrational, and k ≥ ξ √ ! = − π √ , ξ ( √
2) = π √ . Berndt’s work implies that √ j ξ k ( √ j ) π − k ∈ Q whenever j is a positive integer that is nota perfect square, and k ≥ z )with one of the functions { tan( z ) , csc( z ) , sec( z ) } . We can settle the tangent and cosecantcases via elementary trigonometric identities: ∞ X n =1 tan( πnz ) n s = ξ s ( z ) − ξ s (2 z ) , ∞ X n =1 csc( πnz ) n s = ξ s ( z/ − ξ s ( z ) , but it is more challenging to understand the secant zeta function: ψ s ( z ) := ∞ X n =1 sec( πnz ) n s . (1.2) Date : April 5, 2013.2010
Mathematics Subject Classification.
Primary 33E20; Secondary 33B30, 11L03.
Key words and phrases.
Secant zeta function, Bernoulli Numbers, Clausen functions, Riemann zetafunction.This work has been partially supported by NSERC Discovery Grant 355412-2008 and FQRNT Subven-tion ´etablissement de nouveaux chercheurs 144987. The work of FR has also been supported by a Boursed’´et´e de premier cycle du ISM-CRM.
The main goal of this paper is to prove formulas for specials values of ψ s ( z ). In Section2 we prove that the sum converges absolutely if z is an irrational algebraic number and s ≥
2. In Section 4 we obtain results such as ψ ( √
2) = − π , ψ ( √
6) = 2 π . These types of formulas exist because ψ k ( z ) obeys a modular transformation which weprove in Section 3 (see equation (3.8)). Furthermore, based on numerical experiments, weconjecture: Conjecture 1.
Assume that k and j are positive integers, and that k is even. Then ψ k ( √ j ) π − k ∈ Q . The results of Section 4 support this conjecture, even though there are still technicalhurdles to constructing a complete proof. For instance, we prove that the conjecture holdsfor infinite subsequences of natural numbers. The rational numbers that appear are alsointeresting, and we speculate on their properties in the conclusion.2.
Convergence
Since sec( πz ) has poles at the half-integers, it follows that ψ s ( z ) is only well-defined if nz Z + for any integer n . Thus, we exclude rational points with even denominators fromthe domain of ψ s ( z ). If z = p/q with q odd, then ψ s ( p/q ) reduces to linear combinations ofHurwitz zeta functions, and (1.2) converges for s >
1. Convergence questions become morecomplicated if z is irrational. Irrationality guarantees that | sec( πnz ) | 6 = ∞ , but we stillhave to account for how often | sec( πnz ) | is large compared to n s . The Thue-Siegel-RothTheorem gives that | sec( πnz ) | ≪ n ε when z is algebraic and irrational, and this provesthat (1.2) converges for s >
2. The case when s = 2 requires a more subtle argument.We use a theorem of Worley to show that the set of n ’s where | sec( πnz ) | is large is sparseenough to ensure that (1.2) converges. We are grateful to Florian Luca for providing thispart of the proof. In summary, we have the following theorem: Theorem 1.
The series in (1.2) converges absolutely in the following cases: (1)
When z = p/q with q odd and s > . (2) When z is algebraic irrational, and s > . (3) When z is algebraic irrational, and s = 2 .Proof of Theorem 1, parts (1) and (2). Let z be a rational number with odd denominatorin reduced form. It is easy to see that the set of real numbers { sec( nπz ) } n ∈ N is finite. Let M = max n ∈ N | sec( nπz ) | . Then we have | sec( πnz ) | n s ≤ Mn s . It follows easily from the Weierstrass M -test that (1.2) converges absolutely for s > | sec( πnz ) | = | csc ( π ( nz − / | ≪ (cid:12)(cid:12) nz − − k n (cid:12)(cid:12) , (2.1)where k n is the integer which minimizes | nz − − k n | . Now appeal to the Thue-Siegel-RothTheorem [Rot55]. In particular, for any algebraic irrational number α , and given ε > ECANT ZETA FUNCTIONS 3 there exists a constant C ( α, ε ), such that (cid:12)(cid:12)(cid:12)(cid:12) α − pq (cid:12)(cid:12)(cid:12)(cid:12) > C ( α, ε ) q ε . (2.2)If we set α = z , then (2.1) becomes | sec( πnz ) | ≪ n (cid:12)(cid:12) z − k n +12 n (cid:12)(cid:12) ≪ n ε . Therefore we have | sec( nπz ) | n s ≪ n s − − ε , and this implies that (1.2) converges absolutely for s > ε . Since ε is arbitrarily smallthe result follows. (cid:3) In order to prove the third part of Theorem 1, we require some background on continuedfractions. Recall that any irrational number z can be represented as an infinite continuedfraction z = [ a ; a , a , . . . ] , and the convergents are given by [ a ; a , . . . , a ℓ ] = p ℓ q ℓ , which satisfy p ℓ = a ℓ p ℓ − + p ℓ − , (2.3) q ℓ = a ℓ q ℓ − + q ℓ − . (2.4)Convergents provide the best possible approximations to algebraic numbers among rationalnumbers with bounded denominators. In other words, if 0 < q < q ℓ , then (cid:12)(cid:12)(cid:12)(cid:12) z − pq (cid:12)(cid:12)(cid:12)(cid:12) > (cid:12)(cid:12)(cid:12)(cid:12) z − p ℓ q ℓ (cid:12)(cid:12)(cid:12)(cid:12) . (2.5)In addition 1 q ℓ q ℓ +1 > (cid:12)(cid:12)(cid:12)(cid:12) z − p ℓ q ℓ (cid:12)(cid:12)(cid:12)(cid:12) > q ℓ ( q ℓ +1 + q ℓ ) . (2.6)Now we state a weak version of a theorem due to Worley [Wor81, Thm. 1]. Theorem 2 (Worley) . Let z be irrational, k ≥ , and p/q be a rational approximation to z in reduced form for which (cid:12)(cid:12)(cid:12)(cid:12) z − pq (cid:12)(cid:12)(cid:12)(cid:12) < kq . Then either p/q is a convergent p ℓ /q ℓ to z , or pq = ap ℓ + bp ℓ − aq ℓ + bq ℓ − , | a | , | b | < k, where a and b are integers. Now we can complete the proof of Theorem 1. The following proof was kindly providedby Florian Luca.
MATILDE LAL´IN, FRANCIS RODRIGUE, AND MATHEW ROGERS
Proof of Theorem 1, part (3).
Let k n be the integer which minimizes | nz − − k n | . Let W z denote the set of integers where the quantity is large: W z = (cid:26) n ∈ N : (cid:12)(cid:12)(cid:12)(cid:12) nz − − k n (cid:12)(cid:12)(cid:12)(cid:12) ≥ (log n ) n (cid:27) . Then X n ∈ W z | sec( nπz ) | n ≪ | sec( πz ) | + ∞ X n =2 n (log n ) , which converges.Now assume that n W z . Then (cid:12)(cid:12)(cid:12)(cid:12) z − k n n (cid:12)(cid:12)(cid:12)(cid:12) < (log n ) n . Consider the convergents of z . Let ℓ be such that q ℓ − ≤ n < q ℓ . By Theorem 2 thereare at most O (cid:0) (log q ℓ ) (cid:1) solutions to (cid:12)(cid:12)(cid:12) z − p n (cid:12)(cid:12)(cid:12) < (log n ) n (2.7)with p ∈ Z (i.e. consider all values of | a | , | b | < k = 2(log q ℓ ) ).From equations (2.5) and (2.6) we have (cid:12)(cid:12)(cid:12) z − p n (cid:12)(cid:12)(cid:12) ≥ (cid:12)(cid:12)(cid:12)(cid:12) z − p ℓ q ℓ (cid:12)(cid:12)(cid:12)(cid:12) ≥ q ℓ ( q ℓ +1 + q ℓ ) . Combining this with equation (2.4) implies (cid:12)(cid:12)(cid:12)(cid:12) nz − k n (cid:12)(cid:12)(cid:12)(cid:12) n ≥ n q ℓ ( q ℓ +1 + q ℓ ) ≥ q ℓ − q ℓ ( q ℓ ( a ℓ +1 + 1) + q ℓ − ) ≥ q ℓ − q ℓ ( a ℓ +1 + 2) . Hence, if n W z , we find that | sec( nπz ) | n ≪ q ℓ ( a ℓ +1 + 2) q ℓ − ≪ a ℓ +1 q ℓ q ℓ − . Combining the Thue-Siegel-Roth Theorem (equation (2.2)) with (2.6), implies that if z isalgebraic 1 q ℓ q ℓ +1 > (cid:12)(cid:12)(cid:12)(cid:12) z − p ℓ q ℓ (cid:12)(cid:12)(cid:12)(cid:12) > C ( z, ε ) q εℓ . Thus q ℓ +1 ≪ q εℓ . This allows us to place an upper bound on a ℓ +1 : a ℓ +1 ≤ q ℓ +1 q ℓ ≪ q εℓ Putting everything together gives the bound | sec( nπz ) | n ≪ q − εℓ − , and as a result X n W z | sec( nπz ) | n ≪ ∞ X ℓ =1 (log q ℓ ) q − ε ′ ℓ ≪ ∞ X ℓ =1 q − ε ′′ ℓ . ECANT ZETA FUNCTIONS 5
Since q ℓ +1 = a ℓ +1 q ℓ + q ℓ − ≥ q ℓ + q ℓ − , we conclude that q ℓ ≥ F ℓ , where F ℓ denotes the ℓ th Fibonacci number. Since the Fibonacci numbers grow exponentially, we have q ℓ ≫ ϕ ℓ √ , where ϕ = 1 + √ . Setting ˜ ϕ = ϕ − ε ′′ >
1, we finally obtain X n W z | sec( nπz ) | n ≪ ∞ X ℓ =1 ϕ ℓ = 1˜ ϕ − < ∞ . Thus, it follows that (1.2) converges absolutely when s = 2. (cid:3) A modular transformation for ψ k ( z )We begin by noting the following trivial properties of ψ s ( z ): ψ s ( − z ) = ψ s ( z ) , (3.1) ψ s ( z + 2) = ψ s ( z ) , (3.2)2 − s ψ s (2 z ) = ψ s ( z ) + ψ s ( z + 1) . (3.3)The main goal of this section is to prove that ψ k ( z ) also satisfies a modular transforma-tion formula. We start from the partial fractions decompositions of sec( πx ) and csc( πx ),and then perform a convolution trick to obtain an expansion for sec( πx ) csc( πxz ) (equa-tion (3.5)). Differentiating with respect to x then leads to the transformation for ψ k ( z )(equation (3.8)). This method is originally due to the third author, who used used it torediscover the Newberger summation rule for Bessel functions [Rog05], [New82]: J x ( y ) J − x ( y ) π sin( πx ) = ∞ X n = −∞ J n ( y ) n + x . (3.4)Equation (3.4) follows from applying the convolution trick to partial fractions expansionsfor J x ( y ) and J − x ( y ) [Lom68]. Lemma 1.
Let χ − ( n ) denote the Legendre symbol modulo . Suppose that x and z areselected appropriately. Then π csc( πzx ) sec( πx ) = 1 zx + 8 x ∞ X n =1 χ − ( n ) csc ( πnz/ n − x − zx ∞ X n =1 sec ( πn (1 + 1 /z )) n − z x . (3.5) Proof.
Recall the classical partial fractions expansions [GR94]: π (sec( πx ) −
1) =16 x ∞ X n =1 χ − ( n ) n ( n − x ) , (3.6) π csc( πx ) − x =2 x ∞ X k =1 ( − k +1 k − x . (3.7) We assume that z is irrational and algebraic, and x is selected so that the denominators in the sumsare never zero. MATILDE LAL´IN, FRANCIS RODRIGUE, AND MATHEW ROGERS
Both sums converge uniformly. Multiplying the formulas together, expanding via partialfractions, and rearranging the order of summation, we have π (sec( πx ) − (cid:18) π csc( πzx ) − zx (cid:19) =32 zx X n ≥ k ≥ ( − k +1 χ − ( n ) n ( n − x ) ( k − z x )=8 z x X n ≥ k ≥ ( − k +1 χ − ( n ) n ( k − z n / (cid:18) z n / − z x − k − z x (cid:19) =32 zx ∞ X n =1 χ − ( n ) n ( n − x ) ∞ X k =1 ( − k +1 k − z n / ! + 32 zx ∞ X k =1 ( − k +1 k − z x ∞ X n =1 χ − ( n ) n ( n − k /z ) ! . By (3.6) and (3.7) this becomes π (sec( πx ) − (cid:18) π csc( πzx ) − zx (cid:19) =32 x ∞ X n =1 χ − ( n ) n ( n − x ) (cid:18) π csc ( πnz/ − nz (cid:19) + 2 πz x ∞ X k =1 ( − k +1 k ( k − z x ) (sec( πk/z ) − . This reduces to (3.5) after several additional applications of (3.6) and (3.7). We can splitup the sums because all of the individual terms converge absolutely. For instance, we canprove that ∞ X k =1 ( − k +1 sec( πk/z ) k − z x , converges absolutely, by showing that the summand is ≪ | sec( πk/z ) | k − , and then ap-plying Theorem 1 for appropriate choices of z . (cid:3) Theorem 3.
Let E m denote the Euler numbers, and let B m denote the Bernoulli numbers.Suppose that k ∈ N . Then for appropriate choices of z : (1 + z ) k − ψ k (cid:18) z z (cid:19) − (1 − z ) k − ψ k (cid:18) z − z (cid:19) = ( πi ) k k ! k X m =0 (2 m − − B m E k − m (cid:18) km (cid:19) (cid:2) (1 + z ) m − − (1 − z ) m − (cid:3) . (3.8) Proof.
Recall the Taylor series expansions of cosecant and secant: πx csc( πx ) = − ∞ X m =0 (2 m − B m ( πix ) m m ! , sec( πx ) = ∞ X m =0 E m ( πix ) m m ! . Expand both sides of (3.5) in a Taylor series with respect to x . Comparing coefficientsyields the following identity:( πi ) k k ! k X m =0 (2 m − − B m E k − m (cid:18) km (cid:19) z m − = − k ∞ X n =1 χ − ( n ) csc( πnz/ n k + z k − ψ k (1 + 1 /z ) . ECANT ZETA FUNCTIONS 7
Finally, let z (1 + z ) and z (1 − z ), and subtract the two results to recover (3.8). Thecosecant sums vanish because csc( πn (1 + z ) /
2) = csc( πn (1 − z ) /
2) whenever n is odd. (cid:3) We conclude this subsection with a conjecture on unimodular polynomials. We call apolynomial unimodular if its zeros all lie on the unit circle. We have observed numericallythat the polynomials in (3.8) have all of their zeros on the vertical line Re( z ) = 0. Sincethe linear fractional transformation z = (1 − x ) / (1 + x ) maps the vertical line to the unitcircle, we arrive at the following conjecture: Conjecture 2.
We conjecture that the polynomial k X m =0 m (2 m − B m E k − m (cid:18) km (cid:19) ( x − x m )(1 + x ) k − m , (3.9) is unimodular when k ∈ N . This new family of polynomials is closely related to the unimodular polynomials intro-duced in [LR12], [LS13], and [HZ13].4.
Special values of ψ k ( z )Throughout this section we assume that k is a positive even integer. Let SL ( Z ) denotethe set of 2 × ( z ) =SL ( Z ) / h I , − I i . Recall that PSL ( Z ) can be identified with the set of linear fractionaltransformations. The usual group action is (cid:18) a bc d (cid:19) z = az + bcz + d . It is easy to see that multiplying matrices is equivalent to performing compositions oflinear fractional transformations. Consider the following matrices in PSL ( Z ): A = (cid:18) (cid:19) , B = (cid:18) (cid:19) . Equations (3.2) and (3.8) are equivalent to ψ k ( Az ) = ψ k ( z ) , (4.1) ψ k ( Bz ) =(2 z + 1) − k ψ k ( z ) (4.2)+ ( πi ) k k ! k X m =0 (cid:0) m − − (cid:1) B m E k − m (cid:18) km (cid:19) ( z + 1) k − m h (2 z + 1) m − k − (2 z + 1) − k i . Every matrix C ∈ h A, B i has a factorization of the form C = A j B j A j . . . , so equations(4.1) and (4.2) together imply that there exists a z -linear relation between ψ k ( z ), ψ k ( Cz ),and π k .Now we outline a strategy to obtain exact evaluations of ψ k ( z ). First select a matrix C , and then find the linear relation between ψ k ( z ), ψ k ( Cz ), and π k . Next choose z sothat ψ k ( z ) = ψ k ( Cz ). For example, if z = 2 j + p j (2 j + 1) in (4.2), then Bz = z − j MATILDE LAL´IN, FRANCIS RODRIGUE, AND MATHEW ROGERS whenever j is a non-zero integer. With some work the equation collapses to ψ k (cid:16)p j (2 j + 1) (cid:17) = ( πi ) k k ! k X m =0 (cid:0) m − − (cid:1) B m E k − m (cid:18) km (cid:19) × (cid:16) q j j +1 (cid:17) m − − (cid:16) − q j j +1 (cid:17) m − (cid:16) q j j +1 (cid:17) k − − (cid:16) − q j j +1 (cid:17) k − , (4.3)which holds for j ∈ Z \ { } . Similarly, if we take z = | j + 1 | + p j (2 j + 1), then wearrive at ψ k (cid:16) p j (2 j + 1) (cid:17) = ( πi ) k k ! k X m =0 (cid:0) m − − (cid:1) B m E k − m (cid:18) km (cid:19) × (cid:16) q j +12 j (cid:17) m − − (cid:16) − q j +12 j (cid:17) m − (cid:16) q j +12 j (cid:17) k − − (cid:16) − q j +12 j (cid:17) k − , (4.4)which also holds for j ∈ Z \ { } . The right-hand sides of (4.3) and (4.4) are invariantunder the Galois action √ x
7→ −√ x , so both formulas are rational with respect to j .Specializing (4.3) at k = 2 and k = 4 yields ψ (cid:16)p j (2 j + 1) (cid:17) =(3 j + 1) π ,ψ (cid:16)p j (2 j + 1) (cid:17) = (cid:18) j + 46 j + 68 j + 3 (cid:19) π . These results support Conjecture 1. Further evidence for the conjecture is provided bycombining (4.3), (4.4), and (3.3), to obtain identities like ψ (cid:16)p j (2 j + 1) (cid:17) = π , (4.5)whenever j ∈ Z .In general, it seems to be quite difficult to evaluate ψ k ( √ j ) for arbitrary positive integers j . This is due to the fact that (4.1) and (4.2) restrict the available matrices to a subgroupof PSL ( Z ). If there exists a matrix C ∈ h A, B i which satisfies C √ j = √ j , then we canalways evaluate ψ k ( √ j ). We can construct candidate matrices by solving Pell’s equation: X − jY = 1 . If we choose C to be given by C = (cid:18) X j YY X (cid:19) , then it follows that C √ j = X √ j + jYY √ j + X = √ j , and det( C ) = X − jY = 1. Pell’s equationhas infinitely many integral solutions when j ≥ C . The main difficulty is to select appropriate values of X and Y so that C factors into products of A ’s and B ’s. It is not clear if such a selection is always possible.Notice that h A, B i ⊂ Γ (2) and so it follows that PSL ( Z )
6⊂ h
A, B i . ECANT ZETA FUNCTIONS 9
We conclude this subsection by noting that ψ ( z ) = 0 for infinitely many irrationalvalues of z . Setting n = − j in Proposition 1 below, yields ψ r j − ! = 0 , (4.6)for any non-zero integer j . It is worth emphasizing that ψ k ( z ) is highly discontinuous withrespect to z , so these types of results are not surprising. Proposition 1.
Suppose that j and n are integers, and n = 0 . Then ψ r j (2 jn + 1) n ! = (cid:18) jn (cid:19) π . (4.7) Proof.
Setting k = 2 in (4.2) yields ψ ( Bz ) = 12 z + 1 ψ ( z ) + z (3 z + 4 z + 2)(2 z + 1) π . (4.8)Iterating (4.8) gives ψ ( B n z ) = 12 nz + 1 ψ ( z ) + nz (cid:0) z + 4 nz + 2 (cid:1) (2 nz + 1) π . (4.9)The derivation of (4.9) is best accomplished with the aid of a computer algebra systemsuch as Mathematica , because significant telescoping occurs on the right. Now considerthe matrix C = A j B n A j = (cid:18) jn + 1 4 j (2 jn + 1)2 n jn + 1 (cid:19) , and notice that Cz = z , where z = r j (2 jn + 1) n . Thus by (4.9) we have ψ ( z ) = ψ ( Cz )= ψ ( B n ( z + 2 j ))= 12 n ( z + 2 j ) + 1 ψ ( z ) + n ( z + 2 j ) (cid:0) z + 2 j ) + 4 n ( z + 2 j ) + 2 (cid:1) (2 n ( z + 2 j ) + 1) π . We complete the proof by solving for ψ ( z ) and simplifying. (cid:3) Speculations and Conclusion
Assume that k is a positive even integer. Euler gave the following expression forBernoulli numbers: B k = − k !(2 πi ) k ζ ( k ) . (5.1)Bernoulli numbers are interesting combinatorial objects, so it is natural to ask if therational part of ψ k ( √ j ) also has interesting properties. This is an obvious question because ψ k (0) = ζ ( k ). For instance, the von Staudt-Clausen theorem gives a complete descriptionof the denominators of Bernoulli numbers: B k = X ( p − | k p + Integer . (5.2) Is there an analogue of (5.2) for the rational part of ψ k ( √ j )? As an example, consider ψ k ( √
6) which can be calculated from equation (4.3). In order to eliminate some trivialfactors, we define β k := (cid:0) √ (cid:1) k − − (cid:0) − √ (cid:1) k − √ k !( πi ) k ψ k ( √ . By (4.3) we have β k = k X m =0 (cid:0) m − − (cid:1) B m E k − m (cid:18) km (cid:19) k − m (cid:0) √ (cid:1) m − − (cid:0) − √ (cid:1) m − √ . The first few values are β = − / β = 508 /
5, and β = − /
7, which all havedenominators divisible only by primes where p − | k (as hoped for). The first instancewhere this fails is β . The denominator of β equals 5 · ·
11, but 7 − Acknowledgements.
The authors wish to thank Shabnam Ahktari for helpful discussionsand Florian Luca for providing the proof of part (3) of Theorem 1. Finally, the authors aregrateful to the referee for comments and suggestions that greatly improved the expositionof this article.
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E-mail address : [email protected] Department of Mathematics and Statistics, University of Montreal, Montreal, Canada
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