Continued Fractions of Tails of Hypergeometric Series

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1 Continued Fractions of Tails of Hypergeometric Series Jonathan Michael Borwein, Kwok-Kwong Stephen Choi, Wilfried Pigulla. MOTIVATION. The tails of the Taylor series for many stard functions such as the arctangent the logarithm can be expressed as continued fractions in a variety of ways. A surprising side effect is that some of these continued fractions provide a dramatic acceleration for the convergence of the underlying power series. These investigations were motivated by a surprising observation about Gregory s series. Gregory s series for π, truncated at 500,000 terms, gives to forty places 500,000 4 k= k 2k = To one s initial surprise only the underlined digits are wrong i.e, they differ from those of π. This is explained, ex post facto, by setting N equal to one million in the following result: Theorem. For an integer N divisible by 4 the following asymptotic expansion holds: N/2 π 2 2 k= k 2k m=0 E 2m N 2m+ 2 = N N N 5 6 N 7 +, where the numerators,, 5, 6, 385, 5052,... are the Euler numbers E 0, E 2,E 4,E 6,E 8,E 0,... The observation came to the first author s attention in 987. After verifying its truth numerically which is much quicker today, it was an easy matter to generate a large number of the errors to high precision. The authors of [] then recognized the sequence of errors in as the Euler numbers with the help of Sloane s Hbook of Integer Sequences. The presumption that is a form of Euler-Maclaurin summation is now formally verifiable for any fixed N with the aid of MAPLE. This allowed them to determine that is equivalent to a set of identities between Bernoulli Euler numbers that could with considerable effort have been established. Secure in the knowledge that held they found it easier, however, to use the Boole summation formula, which applies directly to alternating series, Euler numbers see []. Because N was a power of ten, the asymptotic expansion was obvious on the computer screen. When one uses N = 0 m in 2 the errors will appear in the mth place until the Euler numbers grow too large. This is a good example of a phenomenon that really does not become apparent without working to reasonably high June July 2005] CONTINUED FRACTIONS OF TAILS OF HYPERGEOMETRIC SERIES 493

2 precision Who recognizes 2, 2, 0? highlights the role of pattern recognition hypothesis validation in experimental mathematics. It was an amusing additional exercise to compute π to 5,000 digits from. Indeed, taking N = 200,000 correcting using the first thous even Euler numbers, Borwein Limber [2] obtained 5,263 digits of π plus 2 guard digits. Thus, while the alternating Gregory series is very slowly convergent, the errors are highly predictable. Although the connection is not immediately obvious, the present paper was stimulated by the third author s attempt to explain the phenomenon via continued fractions. 2. THREE CONTINUED FRACTION CLASSES. In this section we discuss three classes of continued fractions, namely, those of Euler, Gauss, Perron. Euler s continued fraction. Using the following notation for continued fractions, identities such as a a 2 a 3 b ± b 2 ± b = a 3 ± a 2 b ± b 2 ± a 3 b 3 ±..., a 0 + a + a a 2 + a a 2 a 3 + a a 2 a 3 a 4 = a 0 + a are easily verified symbolically. The general form a 2 + a 2 a 0 + a + a a 2 + a a 2 a 3 + +a a 2 a 3 a N a 3 + a 3 a 4 + a 4 = a 0 + a a 2 + a 2 a 3 + a 3 a N + a N 3 can then be obtained by substituting a N + a N a N+ for a N checking that the shape of the right-h side is preserved. This allows many series to be reexpressed as continued fractions. For example, taking a 0 = 0, a = z, a 2 = z 2 /3, a 3 = 3z 2 /5,... using the Taylor expansion arctan z = z z3 3 + z5 5 z7 7 + z9 9, we obtain by letting N in 3 the continued fraction for the arctangent due to Euler: arctan z = z z 2 9z 2 25z z z z. 2 + When z =, this becomes the first infinite continued fraction, given by Lord Brouncker : 4 π = If we let a 0 = N b k be the initial segment of a similar series, we can use 3 to replace the remaining terms with a continued fraction. For example, if we put a 0 = N n z 2n, a = N z 2N+, 2n 2N c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 2

3 then we get a 2 = 2N + 2N + 3 z2, a 3 = 2N + 3 2N + 5 z2,..., arctan z = N n z2n 2n + N z 2N+ 2N N z 2 2N + 5 2N + 3z 2 + 2N + 2 z 2 2N + 3 2N + z 2 2N z N + 7 2N + 5z 2 + Gauss s continued fraction. A rich vein lies in Gauss s continued fraction for the ratio of two hypergeometric functions Fa, b + ; c + ; z/fa, b; c; z see [5]. Recall that within its radius of convergence the Gaussian hypergeometric function is defined by Fa, b; c; z = + ab c z + aa + bb + 2! cc + aa + a + 2bb + b z ! cc + c + 2 The general continued fraction is developed by a reworking of the contiguity relation ac b Fa, b; c; z = Fa, b + ; c + ; z z Fa +, b + ; c + 2; z 7 cc + at least formally is quite easy to derive. Convergence convergence estimates are more delicate. We therefore have Fa, b + ; c + ; z Fa, b; c; z = ac b cc + z 2 Fa +, b + ; c + 2; z z, Fa, b + ; c + ; z which leads to the recursive process for the continued fraction. By taking b = 0 replacing c with c, this process yields in the limit Fa, ; c; z = a z which is the case of present interest. Here a 2l+ = for l = 0,,...We also let a 2 z a + lc + l c + 2l c + 2l, a 2l+2 = F M a, ; c; z = a z a 3 z, 8 l + c a + l c + 2lc + 2l + a 2 z a M z denote the Mth convergent of the continued fraction to Fa, ; c; z. It is well known easy to verify that log + z = z F, ; 2; z. It is a pleasant surprise to discover that log + z z = 2 z2 F2, ; 3; z June July 2005] CONTINUED FRACTIONS OF TAILS OF HYPERGEOMETRIC SERIES 495

4 log + z z + 2 z2 = 3 z3 F3, ; 4; z, then to conjecture that n z n log + z + n = N z N N FN, ; N + ; z. 9 This is straightforward to establish for the first few cases then confirm rigorously. As always, a formula for the logarithm leads correspondingly to one for the arctangent: n z 2 n+ arctan z 2 n + n=0 = N z 2 N+ 2 N + F N + 2, ; N + 32 ; z2. 0 Happily, in both cases 8 is applicable as it is for a variety of other functions such as + z z log, + z k, + t n dt = z F z n, ; + n ; zn. 0 Note that this last function recaptures log + z arctan z for n = 2, respectively. We next give the explicit continued fractions for 9 0. Theorem 2. Gauss s continued fractions for 9 0 are n z n log + z + n = N+ z N N + N 2 z N z N N + 2 z N z N n z 2 n+ arctan z 2 n + n=0 = N z 2N+ 2N + + 2N + 2 z 2 2N z 2 2N N z 2 2N z 2 2N Suppose that we return to Gregory s series but add a few terms of the continued fraction for 0. One observes numerically that, if the results are for N = 500,000, then adding only six terms of the continued fraction has the effect of increasing the precision by forty digits. Example 3. Let N z n E N, M, z := log + z n zn+ N + F MN +, ; N + 2; z 496 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 2

5 E 2 N, M, z := arctan z N n=0 n z 2n+ 2n + + N z 2N+ F M N + 2N + 2, ; N ; z2. Then E N, M, z E 2 N, M, z measure the precision of the approximations to log + z arctan z, respectively, obtained by initially computing the first N terms of their Taylor series then adding M terms of their continued fractions. Tables, 2, 3, 4 record those data for the approximations to log.9,log2,arctan, arctan/2 + arctan/5 + arctan/8, respectively. Note that π 4 = arctan 2 + arctan 5 + arctan is a formula of Machin-type used by Johann Dase in 844 to compute 205 digits of π in his head. 8 Table. Error E N, M, 0.9 for N = 5 0 k k 4 0 M M Table 2. Error E N, M, for N = 5 0 k k 6 0 M M Table 3. Error E 2 N, M, for N = 5 0 k k 6 0 M M June July 2005] CONTINUED FRACTIONS OF TAILS OF HYPERGEOMETRIC SERIES 497

6 Table 4. Error E 2 N +, M, /2 + E 2 N +, M, /5 + E 2 N +, M, /8 for N = 5 0 k k 2 0 M M After some further numerical experimentation it becomes clear that for large a c the continued fraction Fa,, c; z is rapidly convergent. Indeed, the rough rate is apparent. This is part of the content of the next theorem: Theorem 4. Suppose that 2 a, a + c 2a, M 2. If z < 0, then Fa, ; c; z F M a, ; c; z Ɣn + n + aɣn + c aɣaɣc Ɣn + aɣn + caɣc a M 2a c 2, 2 z + 2a c where n =[M/2] F M a, ; c; z is the Mth convergent of the continued fraction to Fa,, c; z. Proof. Recall that Gauss s continued fraction for Fa, ; c; z is where a 2l+ = Fa, ; c; z = a z a 2 z a + lc + l c + 2l c + 2l, a 2l+2 = a 3 z, l + c a + l c + 2lc + 2l + for l = 0,,...Let A n z B n z = a z a 2 z a n z = F n a, ; c; z be the nth convergent of the continued fraction. It can be proved by induction that A z = A 2 z = B z = B 2 z = a z, while when k 3. Hence for k 2wehave A k z = A k z a k za k 2 z, B k z = B k z a k zb k 2 z A k zb k z A k zb k z = a a k z k. 498 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 2

7 Using the estimation in Theorem 8.9 of [3], we find that, if a i > 0foralli,then Fa, ; c; z A nz B n z A n z B n z A n z B n z = a a n z n B n zb n z. It can be shown that the B n z are hypergeometric polynomials see [5]. Explicitly, B 2k z = F k, a k, 2 c 2k; z B 2k+ z = F k, a k, c 2k; z. We then appeal to estimates for hypergeometric polynomials to complete the argument. A more detailed proof can be found in the associated technical report on the CECM preprint server for 2003 at /. In [5] one can find listed many explicit continued fractions that can be derived from either Gauss s continued fraction or its limiting cases. These include the expansions of the exponential function, the hyperbolic tangent, the tangent, various less elementary functions. One especially attractive fraction is that for J n z/j n z I n z/i n z, wherej I are Bessel functions of the first kind. In particular, J n 2z J n 2z = n z z n+ z 2 n+n+2 z 2 n+2n+3. 3 Setting z = i n =, 3 leads to the very beautiful continued fraction I 2 I 0 2 = , since I n z = e nπi/2 J n ze πi/2. In general, arithmetic simple continued fractions correspond to such ratios. An example of a more complicated situation is: 2 z 2 N+ F N +, ; N + 3 ; z N + arcsin z 2 N+2 N+ F, ; = ; z2 σ 2Nz, z 2 where σ 2N is the 2Nth Taylor polynomial for arcsin z/ z 2. Only for N = 0isthis precisely of the form of Gauss s continued fraction. Perron s continued fraction. Another continued fraction expansion is based on Stieltjes work on the moment problem see Perron [4]. In [4, vol. 2, p. 8] one finds the lovely continued fraction z t µ z µ 0 + t dt = z µ + + µ + 2 z µ + 2 µ + z + µ z µ + 3 µ + 2z, + 5 June July 2005] CONTINUED FRACTIONS OF TAILS OF HYPERGEOMETRIC SERIES 499

8 which is valid when µ> < z. We can obtain this as a consequence of Euler s continued fraction if we write z µ z 0 t µ + t dt = z µ + z2 µ z3 µ + 3 z4 µ observe that 5 follows from 3 in the limit here a 0 = 0, a = z/µ +, a n = n+ zµ + n /µ + n for n = 2, 3,... Since z µ+ z F µ +, ; µ + 2; z = µ + 0 t µ dt 6 + t z µ 2 µ+ 2 µ + F + 2, ; µ + 32 z t 2 µ ; z2 = dt t 2 when µ>0, on examining 9 0 we see that this is immediately applicable to provide Euler continued fractions for the tail of the logarithm arctangent series. To be precise, we obtain: Theorem 5. Perron s continued fractions for 9 0 are: n z n log + z + n = N+ z N N 2 z N + N + Nz + n z 2 n+ arctan z 2 n + n=0 = N z 2N+ 2N + + 2N + 2 z 2 2N + 3 2N + z 2 + N + 2 z N + 2 N + z N z 2 2N + 5 2N + 3z +. Moreover, whereas the Gauss Euler/Perron continued fractions obtained are quite distinct, we note the agreement of 9 with 5. Indeed, as we have seen, Theorem 5 coincides with a special case of 3. ACKNOWLEDGMENT. This work was supported by NSERC by the Canada Research Chair Programme. REFERENCES. J. Borwein, P. Borwein, K. Dilcher, Pi, Euler numbers asymptotic expansions, this MONTHLY J. Borwein M. Limber, Maple as a high precision calculator, Maple News Letter ; also available at 3. W. Jones W. Thron, Continued Fractions: Analytic Theory Applications, Encyclopedia of Mathematics Its Applications, vol., Addison-Wesley, Reading, MA, O. Perron, Die Lehre von den Kettenbrüchen, Chelsea, New York, H. Wall, Analytic Theory of Continued Fractions, Chelsea, New York, c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 2

9 JONATHAN BORWEIN was Shrum Professor of Science founding director of the Centre for Experimental Constructive Mathematics at Simon Fraser University In 2004, he re-joined the Faculty of Computer Science at Dalhousie as a Canada Research Chair in Distributed Collaborative Research. Born in St. Andrews in 95, he received his D. Phil. from Oxford in 974, as a Rhodes Scholar. Prior to joining SFU in 993, he worked at Dalhousie 974 9, Carnegie-Mellon , Waterloo His awards include the Chauvenet Prize 993, Fellowship in the Royal Society of Canada 994, in the American Association for the Advancement of Science 2002, foreign membership in the Bulgarian Academy of Sciences Dr. Borwein is governor at large of the MAA a past president of the Canadian Mathematical Society His interests span pure analysis, applied optimization, computational numerical computational analysis mathematics, high performance computing. He has authored ten books most recently two on experimental mathematics a monograph on variational analysis as well as over 250 journal articles. Faculty of Computer Science, Dalhousie University, 6050 University Avenue, Halifax Nova Scotia, Canada B3H W5 jborwein@@cs.dal.ca STEPHEN CHOI was born in Hong Kong obtained his B.Sc. 988 M. Phil. 99 from the University of Hong Kong. Under the supervision of Jeffrey Vaaler, he received his Ph.D. in mathematics from the University of Texas at Austin in 996. In 2000, he joined Simon Fraser University following postdoctoral fellowships at the Institute for Advanced Studies, the Pacific Institute for the Mathematical Sciences UBC SFU, the University of Hong Kong. His research interests lie in studying Diophantine equations, Diophantine approximation, the merit factor of binary sequences. CECM, Department of Mathematics, Simon Fraser University, Burnaby B.C., Canada V5A S6 kkchoi@@cecm.sfu.ca WILFRIED PIGULLA is a retired civil servant amateur mathematician who lives in Germany. June July 2005] CONTINUED FRACTIONS OF TAILS OF HYPERGEOMETRIC SERIES 50

A class of Dirichlet series integrals

A class of Dirichlet series integrals J.M. Borwein July 3, 4 Abstract. We extend a recent Monthly problem to analyse a broad class of Dirichlet series, and illustrate the result in action. In [5] the following