Computing the truncated theta function via Mordell integral

Transcription

1 Computing the truncated theta function via Mordell integral A. Kuznetsov arxiv:36.48v2 [math.nt] 24 Mar 24 Dept. of Mathematics and Statistics York University 47 Keele Street Toronto, ON M3J P3, Canada March 25, 24 Abstract Hiary [3] has presented an algorithm which allows to evaluate the truncated theta function n k= exp2izk +τk2 to within ± in Oln n κ arithmetic operations for any real z and τ. This remarkable result has many applications in Number Theory, in particular it is the crucial element in Hiary s algorithm for computing ζ 2 + it to within ±t λ in O λ t 3 lnt κ arithmetic operations, see [2]. We present a significant simplification of Hiary s algorithm for evaluating the truncated theta function. Our method avoids the use of the Poisson summation formula, and substitutes it with an explicit identity involving the Mordell integral. This results in an algorithm which is efficient, conceptually simple and easy to implement. Keywords: truncated theta function, Mordell integral, Riemann zeta function 2 Mathematics Subect Classification : Y6, M6 Research supported by the Natural Sciences and Engineering Research Council of Canada. The author would like to thank the anonymous referee for careful reading of the paper and for providing many valuable comments and suggestions. The first version of this paper was written while the author was visiting the Department of Mathematical Sciences of the University of Bath, whose hospitality is highly appreciated.

2 Introduction and main results The truncated theta function is defined as F n z, τ := n e 2izk+τk2, k= where n N and z, τ R. We are interested in computing efficiently the truncated theta function F n z, τ for large values of n. The motivation for developing such algorithms comes from computational Number Theory. One particularly important application is for computing the Riemann zeta function ζ + it for large t. The classical result by Riemann and Siegel see formula in [] allows to 2 evaluate ζ + it to within 2 ±t λ in O λ t 2 arithmetic operations by an arithmetic operation we mean additions, multiplications, evaluations of the logarithm and of the complex exponential function. It is a very challenging task to prove that the number of operations can be reduced to Ot α with some α < /2. It is an even harder problem to develop an algorithm which would not only have theoretical complexity Ot α, but would also be feasible for practical implementation. In the recent papers [2, 3], Hiary has developed two new algorithms, which allow us to evaluate ζ + it to within 2 ±t λ in O λ t 3 lnt κ and O λ t 4 3 lnt κ arithmetic operations we will refer to them as /3-algorithm and 4/3-algorithm. The /3-algorithm is particularly attractive compared with the 4/3-algorithm, as it is conceptually simpler, has no substantial storage requirements and should be easier to implement. And this brings us back to the truncated theta functions, as the /3-algorithm is fundamentally based on another result by Hiary see [3, Theorem.], which states that F n z, τ and its derivatives can be computed to within ± in Oln n κ arithmetic operations. Let us summarize the main ideas behind Hiary s truncated theta function algorithm. The algorithm is based on the iterative application of the following two-step procedure. The first step is to normalize z and τ so that they lie in the intervals z [ /2, /2] and τ [ /4, /4]. This is easily achieved via the identities F n z, τ = F n z + 2, τ + 2 = Fn z + k, τ + l, k, l Z, 2 both of which follow directly from. If τ < /n, then we compute F n z, τ by the Euler-Maclaurin summation see Section 3.2 in [3], otherwise we proceed to the second step of the algorithm, which is based on the following identity F n z, τ = e i 4 iz2 2τ F z m, 2τ 2τ 4τ + Rm,n z, τ, 3 where τ > and m = 2nτ. Here and everywhere in this paper x := max{k Z : k x} denotes the floor function. A corresponding identity for τ < can be derived by taking the conugate of both sides of 3 and using the fact that F n z, τ = F n z, τ. Identity 3 plays the central role in Hiary s algorithm. It is derived by applying the Poisson summation formula to and using the crucial self-similarity property of the Gaussian function the fact that taking Fourier transform of exp ax 2 bx c results in a function of the same form but with different parameters. The truncated theta function F m, in the right-hand side of 3 is the dominant contribution arising from the Poisson summation formula, while R m,n z, τ can be considered as the remainder term. The key idea behind the second step in Hiary s algorithm is that the identity 3 transforms a long sum with n terms into a short sum having m = 2n τ n/2 terms recall that we have normalized 2

3 τ so that τ /4. If m is smaller than a fixed power of lnn/, then we compute F m, by direct summation, otherwise we apply the same two-step procedure to F m,. It is clear that this algorithm will terminate after at most log 2 n iterations where log 2 denotes the logarithm with base two. Our goal in this paper is to simplify Hiary s algorithm for computing the truncated theta function. In order to present our results, first we need to introduce Mordell integral, which is defined as hz, τ := R e iτx2 2zx dx, z C, Imτ >. 4 coshx We need to extend this definition for real values of τ. Assume that τ lies in the quadrant Reτ > and Imτ >. Then the following identity is true hz, τ = e i 4 R e τy2 2ze i 4 y coshe i 4 y dy = 2e i 4 i cosh2ze 4 y τy2 e dy. 5 coshe i 4 y This result can be easily established by rotating the contour of integration R + e i 4 R + in the integral 4 and then changing the variable of integration x = e i 4 y. It is clear the integral in 5 provides the analytic continuation of hz, τ into the domain Reτ >, z C. We will adopt this as the definition of hz, τ for τ > and z R, while for τ < and z R we will define hz, τ := hz, τ. Mordell integral and related functions have been studied for more than a century. One particular example of such integrals was used by Riemann to derive the functional equation for the zeta function see section 2. in [], while some further special cases were investigated by Ramanuan [9]. Mordell [6, 7] has developed a general theory of hz, τ and some related integrals. Among many other results, Mordell has discovered the connections between hz, τ and theta functions, he has studied general modular transformations of hz, τ and has proved that hz, τ can be expressed in terms of the Gauss sums when τ is rational more precisely, hz, p/q can be expressed in terms of functions F q z, p/2q and F p z, q/2p, see section 8 in [7]. Recently, the Mordell integral hz, τ was investigated by Zwegers in his Ph.D. thesis [3], and we will follow Zwegers s notation in our paper. The following theorem is our first main result, and it is the basis for our simplified version of Hiary s algorithm for fast evaluations of F n z, τ. Theorem. For τ >, z R and m, n N {} where F n z, τ = e i 4 iz2 2τ F z m, 2τ 2τ 4τ + Rm,n z, τ, 6 R m,n z, τ = i 2 e iz τ 2 h z τ + 2, 2τ 7 i 2 m e 2in+ 2z+τn+ 2 h z + 2n + τ m 2, 2τ. Using Theorem we are able to establish the following result which should be compared with Theorem. in [3]. Theorem 2. There exists an algorithm such that for any,, any z, τ, and any integer n the value of the function F n z, τ can be evaluated to within ± using C ln n 3 arithmetic operations on numbers of C 2 ln n bits. The algorithm requires C3 ln n 2 bits of memory. 3

4 In the above result and everywhere in this paper we assume that A, A 2,... and C, C 2,... are constants, which are positive and absolute, in the sense that they do not depend on the values of any other parameters. The paper is organized as follows: Section 2 contains the proofs of Theorems and 2 and in Section 3 we discuss the practical implementation and some extensions of the algorithm. 2 Proof of the main results Lemma. For z R and τ > hz, τ + hz +, τ = 2 τ e i 4 + i τ z+ 2 2, 8 hz, τ = e i 4 + iz2 τ h z,. 9 τ τ τ Proof. While the above identities are not new see [7] and [3, Proposition.2], we present the sketch of the proof for the sake of completeness. Let us denote θ := e i 4 and assume that τ >. From the second integral representation in 5 we obtain e τy2 2zθy hz + hz + = θ + e 2θy dy = 2θ e τy2 2θyz+ 2 dy. R coshθy R Evaluating the integral in the right-hand side of the above identity use formula in [] gives us 8. In order to prove 9 we use formulas and in [] and evaluate the following two integrals: For w R R e τy2 2zθy 2iyw dy = τ e i τ z+θw2, θ R e 2iyw coshθy dy = cosh θw. Applying Parseval s Theorem for Fourier transform to the first integral in 5 gives us hz, τ = e i τ z+θw2 τ cosh θw dw = e i 4 + iz2 τ hz, = e i 4 + iz2 τ h z,. 2 τ τ τ τ τ R Proof of Theorem : We take the conugate of both sides of equation 8 and obtain Iterating the above identity m times gives us which is equivalent to h z 2, τ = h z + 2, τ + 2 τ e i 4 iz2 τ. 3 h z 2, τ = m+ h z + m + 2, τ + 2 τ m k= k e i 4 i τ z+k2, 4 h z 2, τ = m+ h z + m + 2, τ + 2 τ e i 4 iz2 τ F m 2 z τ, 2τ. 5 4

5 We change variables z = w/t, τ = /t and m = n in 5, and then apply transformation 9 to both h-functions, which results in the following identity e i 4 + i t w t 2 2 h w t 2, t = n+ e i 4 + i t w+nt+ t 2 2 h w + nt + t 2, t 6 + 2e i 4 + iw2 t F n w 2, t 2. At the same time, changing variables z = w + nt m + t/2 /2 and τ = t in 5 we obtain h w + nt m + t, t = m+ h w + nt + t, t e i 4 i t w+nt m+ t F m t t m + w, 2 2t. Eliminating h w + nt + t 2, t from the two equations 6 and 7 gives us e i 4 + i t w 2 t 2 h w t, t = 2e i 4 + iw2 t F 2 n w, t n+ e i 4 + i t w+nt+ 2 t 2 8 m+ [ h w + nt m + t 2, t 2 t e i 4 i t w+nt m+ t F m t m + 2 w, 2t ]. The above identity is equivalent to 6. In order to verify this one would need to apply the transformation F m t m + w, i 2 2t = e t m2 m2w F m t w 2, 2t, 9 which follows easily from by changing the index of summation k m k, then change the variables w = z + and t = 2τ and simplify the result. 2 Now we need to introduce several new obects. We define the sequence {Ẽk} k as coshx = k Ẽ k x k, 2 or, alternatively, Ẽ k = E k /k! where {E k } k are Euler numbers. We define the function f τ z and the sequence of polynomials {q k τ} k as f τ z := e iτz 2 coshz = k q k τz 2k. 2 Using equation 2 it is easy to see that q k τ = k = iτ Ẽ 2k 2. 22! We introduce the sequence of functions {p k x} k, defined by p k x := e xu u k du. 23 5

6 For k, τ > and z R we define where Φz is the error function H k z, τ := e i 4 k [ ] l e i τ z+l+ 2 2 Φ e i 4 τ τ z + l +, 24 2 l= Φz := 2 z Finally, for z > and τ R we define e x2 dx = 2 n n n! z 2n+, z C. 25 2n + where f τ x is given by 2. Jz, τ := e 2xz f τ xdx, 26 Proposition. i For k, z /2 and τ, hz, τ = H k z, τ + H k z, τ + k Jk + z, τ + Jk z, τ. 27 ii For > let us denote K = K = 2+2 ln. Then for any,, z /2 and τ, hz, τ = H K z, τ + H K z, τ + where E <. l K q l τ p 2l 2K + z + p 2l 2K z + E, 28 Proof. Let us prove part i. We denote θ := e i 4, use the second integral representation in 5 and the identity and obtain k coshθy = 2 l e 2l+θy + k e 2kθy coshθy, 29 l= k hz, τ = 4θ l e τy2 2l+θy cosh2zθydy + 2θ k l= e τy2 2kθy cosh2zθy coshθy dy.3 The integrals in the sum in the right-hand side of 3 can be evaluated explicitly by applying formula in []; this gives us the terms H k z, τ + H k z, τ. To deal with the remaining integral in the right-hand side of 3, we rotate the contour of integration R + e i 4 R + and change the variable of integration y = x/θ. This gives us 2θ e τy2 2kθy cosh2zθy coshθy dy = Jk + z, τ + Jk z, τ, 3 6

7 and ends the proof of the identity 27 and part i of the proposition. Let us prove part ii. We write JK + z, τ = I + I 2 = e 2xK+z f τ xdx + e 2xK+z f τ xdx. 32 Using the fact that z /2 and f τ x < for x >, the second integral in the right-hand side of 32 can be bounded from above as I 2 < e x2k dx = e 2K+ 2K < In order to deal with the first integral in the right-hand side of 32, we expand f τ x in Taylor series in x which converges for x < /2, see 2 and obtain I = l q l τ e 2xK+z x 2l dx = l q l τp 2l 2K + z. 34 In order to estimate the tail of the above series, first we will need to establish an upper bound for q k τ. From the following identity for Euler numbers see formulas and in [] Ẽ 2n = E 2n 2n! = 2 n 2n+ 2 k k 2k + 2n+ 35 we conclude that Ẽ2n < 2 2 2n+. Using formula 22 and the fact that τ < we obtain q k τ k Ẽ2k 2! = τ < k 2 = 2k 2+ 2! < 2 2k+ 2k 2 e 2 4 < Using the above result we estimate the tail of the series in 34 as follows q l τp 2l 2K + z < 3 2 2l < l>k l>k Formulas 32, 33, 34 and 37 show that for all z /2 and τ, JK + z, τ = q l τp 2l 2K + z + E, 38 l K where E <. The above statement combined with 27 gives us the desired result Proposition 2. There exists an algorithm such that for any, i and x R the value of Φe 4 x can be evaluated to within ± using A ln arithmetic operations on numbers of A2 ln bits. The algorithm requires A 3 ln bits of memory. 7

8 Proof. Since Φe i 4 x is an odd function, we can restrict x to be a positive number. When x, ] we compute Φe i 4 x using Taylor series 25. Since this series is converging exponentially fast, it can be truncated after Oln terms in order to achieve accuracy ±. Let us consider the case when x,. We will use the following result see equations 3-6 in [4]: For z C with Rez > Φz = hze z2 k= e k2 h 2 + Rz, h + Ez, h, 39 z 2 + k 2 h2 where h > and hrez. Here Rz, h := 2e 2z h The error term Ez, h can be bounded from above as if Rez < h and Rz, h := otherwise. Let us define γ := ln This choice of h implies e 2 h 2 Ez, h, z := e i 4 x and h := 4 and 2 ze z2 e 2 h 2 e 2 2 h 2 Rez2 2 h 2. 4 {, if x 2γ or x 4γ, 2γ, if x 2γ, 4γ. 4 4γ therefore z = Rez2 2 h 2 x 2 2 x 2 h 2 < 4, 42 Ez, h < 84 8 < Next, we define N = 4γ 2 and we estimate the tail of the series in 4 as e k2 h 2 hze z2 z 2 + k 2 h 2 < h khe k2 h 2 < ue u2 du = < 4 2 e Nh2 2 < n N+ n N+ The above results show that for every x > we can choose h according to 4 and obtain i Φ e i he 4 xe ix2 4 x = i N x + 2 e k2 h 2 + E, 45 2 ix 2 + k 2 h 2 where E <. Since the number of terms in the above sum is N = 4 ln, this ends the proof of Proposition 2 in the case x,. Proposition 3. There exists an algorithm such that for any,, z < and τ, the value of hz, τ can be evaluated to within ±/ τ using A 4 ln 2 arithmetic operations on numbers of A 5 ln Nh k= bits. The algorithm requires A6 ln 2 bits of memory. 8

9 Proof. All computations will be performed on numbers of A 3 ln bits, where A3 is the constant from Proposition 2. We set K = ln. The first step is to pre-compute and store in the memory the values of Ẽ k for k 2K. These numbers can be computed recursively via formula 9.63 in [], this computation will require OK 2 arithmetic operations and OK 2 bits of memory. The second step is to use identity 8 and to normalize z so that z /2 note that this will require O z arithmetic operations we will need this fact later. The thid step is to apply formula 28. According to 24 and Proposition 2, the computation of H K ±z, τ to the accuracy of ±/ τ can be achieved in OK 2 arithmetic operations using OK bits of memory. We claim that the computation of the finite sum in 28 to the accuracy ± can also be done in OK 2 arithmetic operations using OK bits of memory provided that we are using the pre-computed values of Ẽ k. This follows from the fact that the coefficients q k τ can be computed via 22 in Ok computations, while the values of p k x can be evaluated recursively via the identity p k x = x kpk x e x, k, 46 which follows easily from 23 by integration by parts. Note that the above recurrence identity is numerically stable as long as x k, which is true in formula 28. Proposition 4. There exists an algorithm such that for any integer n,,, z /2 and τ < n 4 the value of the function F n z, τ can be evaluated to within ± using A 7 ln n 3 arithmetic operations on numbers of A 8 ln n bits. The algorithm requires A9 ln n 2 bits of memory. Proof. The main idea behind this algorithm is to expand the exponential function in Taylor series, however the details of the implementation will be different depending on whether z > n or z < n. Let us consider the first case, when z > n. We expand exp2iτk 2 in Taylor series and obtain [ n ] 2i l F n z, τ = τk 2 l e 2izk. 47 l! l= Since τ < n 4, the absolute value of the sum in the square brackets is bounded from above by n +. Therefore, the external sum in l is converging exponentially fast, and in order to achieve accuracy ± we can truncate it after L A ln n terms for some constant A. We assume that L 3 < n, otherwise we will compute F n z, τ by direct summation. Our goal is to show that the sum in the square brackets can be evaluated with accuracy ± in OL 2 operations on numbers of OL bits using OL 2 bits of memory. The main idea is to rewrite this sum as follows n τk 2 l e 2izk = k= Using Leibniz rule we find that 4l d2l n dz 2l τ l d2l 2i 2l dz 2l n k= k= [ ] e 2izn+ 2l = e 2iz e 2izk = n τn4 l d2l n 4l 2i 2l dz 2l = [ e 2izn+ e 2iz ]. 48 2l f zg 2l z, 49 where we have defined [ 2 d f z := n dz e 2iz ], 5 9

10 and 2 d [ g z := n e 2izn+ ] = n 2 2in + e 2izn+ dz {=}. 5 While the computation of g z is straighforward, the computation of f z requires more work. First, we check by induction that + f z = a,k nexp2iz k, 52 k= where a, = and the remaining coefficients a,k can be computed by the recursion a +,k = 2i n k a,k + kn a,k {k +},, k From 53 one can see by induction that a,k < 4/n < recall that < 2L < 2n 3. Note that the condition n < z /2 implies n exp2iz > maxn sin2z, n cos2z 4, 54 since sin2z 4 z if z /4 and cos2z > if /4 < z < /2. Given 54 and the above upper bound on the coefficients a,k, it is clear that formula 52 provides a numerically stable way of computing f z using numbers of OL bits. The memory requirement is OL 2 bits, since in order to compute the coefficients a +,k, k + 2 via 53 we need to store at most 2L + numbers a,k, k +. When z < n the lower bound 54 is no longer valid, and we have to proceed by a different route. In this case we expand the exponential function exp2izk + τk 2 in Taylor series and obtain F n z = [ n ] 2i l zk + τk 2 l. 55 l! l k= Since z < n and τ n 4, the sum in the square brackets is bounded from above by n + 2 l. Therefore, the sum in l is converging exponentially fast, and in order to achieve accuracy ± we can truncate it after L = O ln n terms. The sum in the square brackets in 55 can be computed as follows n l l zk + τk 2 l = nz l n 2 τ S l+ n, 56 k= where we have defined S n := n n k= k. Formula in [] gives us S n = n + i= = + i B i n i,, n, 57 i where B i are the Bernoulli numbers {, /2, /6,... }. We will leave it to the reader to verify that the above three formulas provide the required algorithm for computing F n z, τ to within ± in OL 3 arithmetic operations on numbers of OL bits one should precompute and store 2L values of B i /i!, i 2L, which can be done in OL 2 arithmetic operations using OL 2 bits of memory.

11 Proof of Theorem 2: We are given z,, τ,, a positive integer n and a small positive number. We will describe the algorithm for computing the value of F = F n z, τ. In order to start the algorithm, we use identities 2 and normalize z and τ so that z /2 and τ /4; we define z and τ to be equal to these normalized values of z and τ. The algorithm is based on a recursion, and will be the counter which keeps track of the steps of the recursion. We initialize =, n = n, α = and β =. All computations are performed on numbers of max5a 5, 3A 8 ln n bits, where A5 and A 8 are the constants appearing in Propositions 3 and 4. The algorithm: i If n lnn 3 then we compute f = F n z, τ by direct summation. Terminate the algorithm and return F = α f + β. ii If τ < n 4 then we compute f = F n z, τ to the accuracy of ±/n 3 using the algorithm from Proposition 4. Terminate the algorithm and return F = α f + β. iii If τ n 4 we set n + = 2n τ, z = z 2 τ values of z and τ use identities 2. If τ >, then α + = while if τ < then α + = and τ = 4τ. Set z + and τ + to be the normalized α i exp 2τ 4 iz2, β + = β + α R n+,n τ z, τ, 58 α 2 τ exp i 4 iz2, β + = β + α R n+,n τ z, τ, 59 where R m,n z, τ is given by by 7, and the values of the Mordell integral h, appearing in 7 are computed using the algorithm from Proposition 3 to the accuracy ±/n 3. iv Increase the counter + and proceed to step i. Assume that this algorithm stops after J iterations. The fact that this algorithm returns the correct value of F n z, τ can be verified by induction on J using identity 6. Let us investigate the number of arithmetic operations required by this algorithm. At each iteration of the algorithm, provided that it does not terminate in steps i or ii, we have the new value n + which satisfies n + = 2n τ n /2 recall that τ /4. This shows that the algorithm will either terminate in step i after at most log 2 n iterations, or it will terminate in step ii before that. Let us denote L = ln n. Step ii resp. step iii requires OL 3 resp. OL 2 arithmetic operations and both of these steps require OL 2 bits of memory see Propositions 3 and 4. Since step ii will be executed at most once, and step iii at most log 2 n times, it is clear that the algorithm requires OL 3 arithmetic operations and OL 2 bits of memory. Finally, let us consider the accuracy of this algorithm. All numbers appearing in the algorithm are evaluated to the accuracy ±/n 3. Since the algorithm did not terminate at the iteration J, we have n J > and τ J n 4 J > n 4. Recall that for all we have n + 2n τ, therefore J 2 < n J 2 J 2 n τ i, 6 i=

12 and applying formulas 58 and 59 we obtain α J = [ 2 J 2 J 2 τ i i= ] 2 < n. 6 Assuming that the algorithm terminates in step ii, then the final accuracy is at least ±/n 3 n log 2 n, which is smaller than the required accuracy ±. On the other hand, if the algorithm terminates in step i, then α J = α J / 2 τ J < n 5 2, and the final accuracy is ±/n 3 n 5 2 log 2 n, which is still smaller than ±. Remark : One may ask the following natural question: is the choice n + = 2n τ in the above algorithm optimal? In other words, given that the identity 6 is true for all positive m and n, why can not we choose m = n + 2n τ, thus reducing the number of terms in the new sum F m,? The rationale for this choice is that the computation R m,n, in formula 7 requires the evaluation of the Mordell integral hz + 2n + τ m 2, 2τ, which involves more than 2n τ m arithmetic operations see step 2 in the proof of Proposition 3. Therefore, while the choice of m = n + 2n τ will decrease the number of terms and the computation time of F m,, any gain will be canceled by the corresponding increase in the computation time of R m,n,. 3 Practical implementation and extensions of the algorithm As is often the case, the algorithm which can be analyzed analytically and which allows for rigorous error bounds is not necessarily the most efficient algorithm from the practical point of view. While it is certainly possible to perform practical computations of F n z, τ using the algorithm described in the proof of Theorem 2, in this section we will explain how one could produce a more efficient algorithm with a certain amount of pre-computation and a few numerical experiments. This practical implementation is suitable in the case when we need to compute F n z, τ to a fixed accuracy ± for many different values of z, τ and n N for some fixed value of N. As we saw in the proof of Theorem 2, the main computational effort is spent in evaluating F n z, τ for very small values of τ when τ < n 4 and in computing the Mordell integral hz, τ. We do not see a way of making the former of these tasks much faster, however the latter task can certainly be done much more efficiently. Part i of Proposition shows that in order to compute hz, τ we need to be able to evaluate the error function Φe i 4 x for x R and to compute Jz, τ defined by 26. Let us first discuss the computation of the error function. Our approach to computing Φe i 4 x is to divide the interval, into a number of sub-intervals < x < x 2 < < x m = x <, and use the Chebyshev approximation on each sub-interval. On the infinite interval x, we define the function fσ via Φe i 4 x = + e ixz2 x fσ, where σ := x /x 2 and we approximate fσ by the first few terms of the Chebyshev series. It is known see [8] that the coefficients of the corresponding Chebyshev series decay as exp 2 nx On 2, therefore, by a proper choice of x we can be sure that we need only a few terms of the Chebyshev series to obtain the required accuracy. Once we have chosen x, we divide, x into m sub-intervals of equal length, and on each of them we compute the first few terms of the Chebyshev series. Note that on each finite interval x i, x i+, i < m, the Chebyshev series approximating Φe i 4 x must converge exponentially fast since Φz is an entire functions. By choosing 2

13 m large enough we can make sure that the number of significant terms in each Chebyshev series is small. For example, in our implementation of this algorithm we chose x = 5 and m = 5, and on each subinterval we approximated Φe i 4 x by the first thirty terms of Chebyshev series. This approximation had absolute error 3 over all real values of x. The second important problem is how to evaluate Jk + z, τ efficiently. Our approach is based on the following formula Jk + z, τ = 2k e y g z,τ y k dy, where gz,τ y := e iτy2 4 zy cosh y 62, 2 which follows from 26 by changing the variable of integration x = y/2k. For k, z /2 and τ, the function y, g y z,τ k is bounded, and as k + it converges to gz,τ =. Therefore, when k is reasonably large, the integral in 62 can be computed to a very high-precision using the Gauss-Laguerre quadrature with the weight function e y and M nodes. The problem is to decide what reasonably large means, and here one should do a number of numerical experiments to find the optimal values of k and M. If we take k to be a large number, then the function g y z,τ k is very close to, and M the number of nodes in Gauss-Laguerre quadrature can be taken quite small in order to achieve the required accuracy. Of course, the disdvantage of choosing k to be large is that it would require many evaluations of the error function in 24, and it would increase the run-time of the algorithm. On the other hand, if we take k to be a small integer, then the function g y z,τ k oscillates and M has to be very large in order to provide the required accuracy, which would again increase the run-time of the algorithm. Therefore, k and M have to be chosen so that the computation time of H k z, τ it approximately equal to the computation time of Jk + z, τ. In our examples we took k = 5 in formulas 27, 24 and 26 and we have used the Gauss-Laguerre quadrature with M = 24 nodes truncated to the smallest 4 nodes, see [5] to evaluate the integral in 26. In order to verify the accuracy, we have computed hz, τ on a very fine regular grid of points in the rectangle z /2 and τ, /2 using the above algorithm and we have checked that the relative error is always less than 29 when compared with the algorithm described in Proposition 3. While the algorithm based on Gauss-Laguerre quadrature is very efficient, we were not able to provide rigorous error estimates. It is known that the error of the M-point Gauss-Laguerre quadrature can be bounded by a multiple of { } 2M η 2M := sup y g 2M z,τ y : y, 63 see Theorem 3 in [], but we were not able to obtain good upper bounds for this quantity. The results of our numerical experiments are presented in Figure. The algorithm was implemented in Fortran using the quadruple precision. Due to the fact that the number of iterations of the algorithm varies for different values of τ, we have tested the algorithm for random pairs z, τ, sampled uniformly from the domain < τ < /4 and /2 < z < /2, and the results in Figure represent the average run-time for a single evaluation of F n z, τ. Since we are working in fixed precision which does not depend on n, as in the algorithm in the proof of Theorem 2, these computations involve an unavoidable loss of precision, which becomes more pronounced as n increases. See the paragraph preceding Remark on page 2, explaining the reason for this loss of precision. Our results indicate that the difference between the values of F n z, τ computed via direct summation and our version of Hiary s algorithm is typically of the order of 28 for n = 3 and around 25 for n = 5. This loss of precision is still acceptable for practical purposes. Most importantly, the figures b and d confirm 3

14 a b c d Figure : Average run time for computing a single value of F n z, τ x-axis represents n, y-axis represents time in seconds. The curve marked with circles corresponds to direct summation algorithm based on, the curve without markers corresponds to Hiary s algorithm. that the run-time of the algorithm increases essentially logarithmically with n, and that the simplified version of Hiary s algorithm is much faster than the summation of n terms in even for moderately large n. Finally, we will discuss a related problem of evaluating the finite sum of the form F n, z, τ := n n k= k e 2izk+τk2 = 2in z F nz, τ. 64 Hiary s result [3, Theorem.] states that F n, z, τ can be computed to the required accuracy ± in poly-log time in n/ though the precise statement is slightly more complicated, as the implied constants also depend on. Our results can also be adapted to give a simpler version of Hiary s poly-log time algorithm for computing F n, z, τ. Below we will sketch the main ideas of the practical implementation of such an algorithm. We will follow the same path as in the algorithm described in the proof of Theorem 2. Our goal is to compute the values of F n, z, τ for J. At each step of the recursion, we normalize the values of z and τ so that z /2 and τ /4. If τ < n 4, then we compute F n, z, τ by expanding the exponential function in 64 in Taylor series and applying similar ideas as in the proof of Proposition 4. 4

15 If τ n 4, then we use the following identities i 4 z F nz, τ = e 2τ k= [ ] [ k iz 2 k e 2τ k z k z F z k m, ] + 2τ 4τ z R m,nz, τ, J, 65 which follow from 6 by applying Leibniz rule. These identities reduce the computation of F n, z, τ to the computation of F m,, with m = 2nτ < n/2 and J, and complete the main step of the recursion. Applying identities 65 in practice will involve computing hz, τ, which is equivalent to evaluating z G = H z k z, τ and G 2 = Jk + z, τ see formulas 24, 26 and 27. Computing G z does not pose a serious problem, as applying Leibniz rule to 24 would give us an explicit expression for G, and since Φ z = 2/ exp z 2 it is easy to see that this explicit expression would involve only elementary functions and the error function Φz. At the same time, when z is large it will be more efficient to compute e z2 Φz directly by taking derivatives of the right-hand side of formula z 45 or by using the asymptotic expansion for this function, see formula in []. The value of G 2 = Jk + z, τ can be computed using the generalized Gauss-Laguerre quadrature. Indeed, from z 26 we find that Jk + z, τ = z 2 k e y y g y z,τ k dy, 66 where g z,τ y is defined in 62. Using the same strategy as discussed on page 3 following equation 62 the integral in the right-hand side of 66 can be evaluated using the generalized Gauss-Laguerre quadrature with the weight function x e x and M nodes. By experimenting with different values of k and M and choosing the optimal ones we can compute hz, τ very efficiently with the required z accuracy. References [] I. S. Gradshteyn and I. M. Ryzhik. Table of integrals, series, and products. Elsevier/Academic Press, Amsterdam, seventh edition, 27. [2] G. A. Hiary. Fast methods to compute the Riemann zeta function. Ann. of Math., 742:89 946, 2. [3] G. A. Hiary. A nearly-optimal method to compute the truncated theta function, its derivatives, and integrals. Ann. of Math., 742: , 2. [4] D. Hunter and T. Regan. A note on the evaluation of the complementary error function. Math. Comp., 268:539 54, 972. [5] G. Mastroianni and G. Monegato. Some new applications of truncated Gauss-Laguerre quadrature formulas. Numerical Algorithms, 49-4: , 28. [6] L. J. Mordell. The value of the definite integral 342, 92. e at2 +bt dt. Quarterly Journal of Math., 68:329 e ct +d 5

16 [7] L. J. Mordell. The definite integral 6:322 36, 933. e at2 +bt dt and the analytic theory of numbers. Acta Math., e ct +d [8] G. Nemeth. Chebyshev expansions for Fresnel integrals. Numerische Mathematik, 7:32 32, 965. [9] S. Ramanuan. Some definite integrals connected with Gauss sums. Messenger of Mathematics, 44:75 85, 95. [] A. H. Stroud and K. W. Chen. Peano error estimates for Gauss-Laguerre quadrature formulas. SIAM J. Numer. Anal., 92: , 972. [] E. Titchmarsh. The theory of the Riemann zeta-function. Oxford University Press, second edition, revised by D. R. Heath-Brown, 986. [2] W. Van Snyder. Algorithm 723: Fresnel integrals. ACM Trans. Math. Softw., 94: , Dec [3] S. Zwegers. Mock theta functions. Ph.D. thesis, Utrecht University, arxiv: , 22. 6