RAMANUJAN: A TALE OF TWO EVALUATIONS

Transcription

1 ROCKY MOUNTAIN JOURNAL OF MATHEMATICS Volume 46, Number 3, 06 RAMANUJAN: A TALE OF TWO EVALUATIONS DONALD J. MANZOLI ABSTRACT. In 887, beneath a canopy of stars, Srinivasa Ramanujan commenced his brief eistence on this planet. In a universe at the mercy of its entropy, against all odds, a genius was born. His destiny was mathematics, a subject born thousands of years earlier. The power of this discipline is not to be denied. After all, with our minds in the stars, we have placed footprints on the moon. The conquest of the moon was a triumph of applied mathematics; however, it was the landscape of pure mathematics that awaited Ramanujan. In time, he would eplore it with passion, leaving footprints lasting for eternity. Professor Bruce C. Berndt has done a remarkable job of editing the notebooks Ramanujan left behind. In particular, Berndt s Chapter 9 of Ramanujan s notebooks Part I provides a magnificent in-depth look at raw mathematical talent in action. The primary purpose of this article is to present three Chapter 9 related results, a series evaluation and two new functional equations, that Ramanujan either missed or his work on them was lost. The secondary purpose is to present what Berndt calls a corrected version [5, page 33] of an incorrect Chapter 9 formula of Ramanujan. This series evaluation represents one of the few serious mistakes to be found in Ramanujan s work.. A missing functional equation. This mathematical pioneer had a marvelous way of looking beyond the familiar, using his intuition to guide him through what, in his mind, was uncharted mathematical territory. For eample, consider the following well-known power series: where f) = H k = k j= H k k+ k + j = ln ), and <. 00 AMS Mathematics subject classification. Primary 40G99. Keywords and phrases. Ramanujan, polylogarithms. Received by the editors on August 8, 0, and in revised form on May, 04. DOI:0.6/RMJ Copyright c 06 Rocky Mountain Mathematics Consortium 95

2 96 DONALD J. MANZOLI For = /, this yields ) H k f = k + ) k+ = ln. Ramanujan considered two analogues of f), namely, H k k+ g) = k + ), h) = H k k+ k + ) 3, where [5, pages 5, 53]. For simplicity in what follows, we will consider only real arguments.) Ramanujan s entry 9 i) [5, page 5], [9, page 303, )] provides a formula, valid for 0 < <, that allows g/) to be evaluated..) g ) = ln ln ) + Li ) ln + ζ3) Li 3 ), where Li n ) = k, for n > and, kn denotes the nth polylogarithm function [5, page 46]. Equivalently,.) g ) = ln ln ) + ζ) ln Li ) ln + ζ3) Li 3 ). Since Li /) = /) ln + /)ζ) and Li 3 /) = /6) ln 3 /)ζ) ln + 7/8)ζ3), we have ) H k g = k + ) k+ = 6 ln3 + 8 ζ3). Having evaluated f/) and g/), it is reasonable to investigate the possible evaluation of ) H k h = k + ) 3 k+. Although Ramanujan developed functional equations for g ) g /)) and h ) h /)) entry 9 ii) [5, page 5] and entry 0 i) [5, page 53], respectively), he evidently did not establish

3 RAMANUJAN: A TALE OF TWO EVALUATIONS 97 a formula for h ) alone, analogous to.) for g ) entry 9 i)). Lewin [9] does not list such a result for h ) either. The missing formula, developed by the author, is h ) = 4 ln4 6 ln3 ln ) + ζ) ln.3) + ζ3) ln Li 3 ) ln + ζ4) Li 4 ) + Li 4 ) + Li 4 ), where / <. For = /, this yields ) H k.4) h = k + ) 3 k+ = 4 ln4 8 ζ3) ln + 8 ζ4). Note the similarity of.3) and.), which is obtained from.) by using a functional equation for the dilogarithm attributed to Euler Ramanujan s entry 6 iii) [5, page 47])..5) Li ) + Li ) = ζ) ln ln ), where 0 < <. Note also the cancellation of Li 4 /) in the evaluation of h/). Although Li /) = /) ln +/)ζ) and Li 3 /) = /6) ln 3 /)ζ) ln + 7/8)ζ3), a similar epression for Li 4 /) is not known [0, page ]. The proof of.3) requires bringing in one additional formula, which is due to Landen Ramanujan s entry 7 i) [5, page 49]). Li 3 ) + Li 3 ) + Li 3 ).6) = 6 ln3 ln ln ) + ζ) ln + ζ3), where / <. Solving.) for Li 3 ) Li ) ln and adding it to.6) gives us a new functional equation for the trilogarithm..7) Li 3 ) + Li 3 ) Li ) ln = 6 ln3 + g ), where / <. This can be derived directly for /.) The proof of.3) involves creating applications of.6) and.7). For

4 98 DONALD J. MANZOLI / <, we have Li 4 )) + Li 4 )) Li 4 )) Li 3 ) ln ) ) k ) /)) k ) = + k 4 k k 4 ) ln k 4 ) k = Li 3 ) + /) Li 3 ) Li 3) Li 3 ) Li ) ln = Li 3 ) + Li 3 ) ) Li ) ln + Li 3 ) + Li 3 ) ) + Li 3 ) = ) 6 ln3 + g ) + 6 ln3 ) ln ln ) + ζ) ln + ζ3) = ln ln ζ) ln ζ3) k 3 ) ln ) ln 3 6 g ). Integrating, we arrive at a new functional equation for the quadrilogarithm. Li 4 ) + Li 4 ).8) Li 4 ) Li 3 ) ln = 4 ln4 + 6 ln3 ln ) ζ) ln ζ3) ln ζ4) + h ), where / <. Solving this for h ) gives us.3).

5 RAMANUJAN: A TALE OF TWO EVALUATIONS 99 It is possible to evaluate h/) without utilizing.3). To do this, we need Ramanujan s entry 0 i) [5, page 53]. h ) h ) = 4 ln4 + 6 ln3 ln ) + ζ3) ln + Li 3 ) ln + ζ4) Li 4 ), where / <. For = /, this yields ) ).9) h + Li 4 h ) = 4 ln4 + ζ) ln 5 ζ3) ln + ζ4). 8 A formula for h ) given by Borwein, Borwein and Girgensohn [7, page 6] can be restated as.0) ) Li 4 h ) = ln4 + ζ) ln 7 5 ζ3) ln ζ4). Subtracting.0) from.9), we again obtain h/) = /4) ln 4 /8)ζ3) ln + /8)ζ4), which is.4).. A corrected version. As pointed out in the Frontline article [6], Rediscovering Ramanujan, Most of the time Ramanujan provided only the results and not the proof. Thus, Berndt s editing job is doubly difficult. He is interviewed in this article, and one of the things he said especially caught our attention, I did not count the serious mistakes but it is an etremely small number maybe five or ten out of over 3,000 results. Considering that Ramanujan did not have any rigorous training, it is really amazing that he made so few mistakes. The secondary motivation for writing this article is to present what Berndt calls a corrected version [5, page 33] of an incorrect formula of Ramanujan. The identity of Ramanujan we address appears in Berndt s Ramanujan s notebooks, Part I. There he eplains that, Unfortunately, this beautiful formula is incorrect. He goes on to say that,

6 930 DONALD J. MANZOLI We have been unable to find any formula for G) which resembles.3) [5, page 57]. Restated slightly, the formula is.) where k) 3 = π 4 ) k+ 4k 3) 3 π 3 3 = k j= j. k ) 3, Although this is an attractive evaluation, it misses the mark. thousand terms of k) 3 yields approimately 0.6. A thousand terms of each sum on the right hand side of the identity yields approimately 0.44 for Ramanujan s epression. Defining λn) = k ) n for n >, see [, page 807]), and utilizing Euler s identity found in [4, page ],.) = π 3 3, we can further restate.) as.3) k) 3 = π 4 ) k+ 4k 3) 3 λ3) 3k )3k ). Unepectedly, we found a formula bearing a strong resemblance to.3)..4) k ) 3 + k) 3 = π 6 + λ3) ) k+ k ) 3 k 0)k ). A

7 RAMANUJAN: A TALE OF TWO EVALUATIONS 93 By comparison, a thousand terms of each sum in this identity yields approimately.36 for each side. The proof of.4) consists of developing the following three identities.5) j= k j )j + k) =,.6) j= k)j ) j + k) = 4 ζ)λ) k) 3,.7) j= k)j ) j + k) = η)λ3) + where ηn) is defined to be ) k+ k n for n see [, page 807]). Note that where βn) is defined to be 4 ζ)λ) = λ4) = 3 β)β3), ) k+ k ) n for n k ) 3, see [, page 807]). The proof of.5) is simple and is omitted here. Each double sum identity proof requires a different initial algebraic trick followed by an application of.5) at the end of the argument. First, we have j= = j= k)j ) j + k) k + j ) j ) k) j ) j + k)

8 93 DONALD J. MANZOLI = j= j= k) j ) = 4 ζ)λ) = 4 ζ)λ) k) j )j + k) j= k) 3. k k) 3 j )j + k) This completes the proof of.6). Similarly, we have j= = = k)j ) j + k) j= j= = + j= k )j ) k)k )j ) 3 j + k) j + k) + k)j) k)k )j ) 3 j + k) j= = η)λ3) + = η)λ3) + = η)λ3) + k)k )j ) 3 j k )j ) 3 j + k) j= j= This completes the proof of.7). j k )j ) 3 j + k) k k ) 3 j )j + k) k ) 3.

9 RAMANUJAN: A TALE OF TWO EVALUATIONS 933 Identities.6) and.7) give us 4 ζ)λ) which simplifies to.4) since k) 3 = η)λ3) + k ) 3, 4 ζ)λ) = 3 β)β3) and β) = π 4. This completes the proof of.4), our corrected version of an incorrect formula of Ramanujan. 3. Intriguing comparisons. The elementary techniques used in this paper can by lengthy arguments) be used to derive the first and third entries of the following two identity sets. Since 3 β)β3) = λ4),.4) can be involved in two additional intriguing comparisons. 3.) k) = λ3) k ) 3 + k) 3 = λ4) + η)λ3) k) 4 = λ5) ξ)λ3) k ) = λ3) + η)λ) k) 3 + k ) 3 = λ4) + η)λ3) k ) 4 = λ5) + η)λ4) λ)ξ3),

10 934 DONALD J. MANZOLI where ξn) is defined to be Thus, k) n for n >. k ) 3 + k) 3 is, in a sense, behaving as one might epect to behave alone. k) 3 or k ) 3 A quick comment on a correct result of Ramanujan is in order here. After constructing a proof of the formula k)k ) = ζ), we were surprised to find a formula of Ramanujan see [5, page 50]) that can be used in conjunction with the power series for arctanh ) to create a different proof. We did not think it was possible to prove it in any other way, but Ramanujan proved us wrong. Let us define m = k j= for m. j ) m For m =, this leads to a third double sum identity that allows us to say something new about j= k) 3. k)j ) j + k) = 8 λ4) + k). Since 4 ζ)λ) = λ4),

11 RAMANUJAN: A TALE OF TWO EVALUATIONS 935 we can then recall.6) j= k)j ) j + k) = λ4) and write a new identity reminiscent of which is equivalent to 3.). 3.) k It can also be shown that 3 k + Unlike k k = λ3), + k 3 and k 3 = 3λ4). + k 4 = 4λ5). k 3, k) 3 all three of these sums can be readily evaluated in terms of λn) and ζn). It is also true that 4 k + 3 k 3 + k 4 + k 5 = 5λ6). We suspect that k 5 A general formula for will share the same fate as k. 3 k) n eists and can be found in [8, page 68]. Thus, a satisfactory evaluation of k) 3

12 936 DONALD J. MANZOLI is an elusive entity; however, it is straightforward to verify that H k k 3 = ζ ). Using elementary methods that rely on partial sum identities, we also proved the more difficult formula H k k 3 = 7 ζ5) ζ)ζ3). Later we learned that this can be derived by more advanced methods see [3, page 4]). The computational tools employed in these methods are something that Ramanujan did not have available to him. Computer-aided symbolic manipulation [, page 8] and computerized integer relation detection algorithms [, page 3] are now a reality. Ramanujan s path to mathematical truth was not digital, but it must have been delightful. It should be mentioned that.4) is the second of four similar formulas that we proved in basically the same way. 3.3) 3.4) k ) k) = η)λ) k ) 3 + k) 3 = λ4) + η)λ3) k ) 4 = η)λ4) + ξ)λ3) λ)ξ3) k) 4 k ) 5 + k) 5 = λ6) + η)λ5) λ3)ξ3). The last identity can be used to enhance the two identity sets introduced earlier. k) = λ3) k ) 3 + k) 3 = λ4) + η)λ3)

13 RAMANUJAN: A TALE OF TWO EVALUATIONS 937 Now both and k) 4 = λ5) ξ)λ3) k ) 5 + k) 5 = λ6) ξ3)λ3) + η)λ5) k ) = λ3) + η)λ) k) 3 + k ) 3 = λ4) + η)λ3) k ) 4 = λ5) + η)λ4) λ)ξ3) k) 5 + k ) 5 = λ6) + η)λ5) λ3)ξ3). k ) 3 + k) 3, k ) 5 + k) 5, are, in a sense, behaving as one might epect the component sums to behave alone. Identity 3.3) and its successor 3.4) may be as close as one can get to the kind of evaluation Ramanujan was seeking for k) A conjecture. Finally, recall from 3.) that k) + k) 3 = 3 8 λ4). Clearly, k) shares the same fate as k), when it comes 3 to being evaluated. With this in mind, we conclude with the following

14 938 DONALD J. MANZOLI conjecture. n k) = n )n)λn + ) 4 k)λk + )λn k). n Acknowledgments. I wish to thank Professor Paul Ezust of Suffolk University for encouragement. REFERENCES. M. Abramowitz and I.A. Stegun, eds., Handbook of mathematical functions, Dover, New York, D.H. Bailey and J.M. Borwein, Eperimental mathematics: Recent developments and future outlook, preprint, 999, available at carma.newcastle.edu.au/50/. 3. D.H. Bailey, J.M. Borwein and R. Girgensohn, Eperimental evaluation of Euler sums, preprint, 993, available at newcastle.edu.au/60/. 4. L. Berggren, J.M. Borwein and P.B. Borwein, Pi: A source book, Springer- Verlag, New York, B.C. Berndt, Ramanujan s notebooks Part I, Springer-Verlag, New York, , Rediscovering Ramanujan, Frontline 6, 999, available at hindu.com/fline/fl67/67080.htm. 7. D. Borwein, J.M. Borwein and R. Girgensohn, Eplicit evaluation of Euler sums, preprint, 993, available at edu.au/58/. 8. P.F. Jordan, Infinite sums of Psi functions, Bull. Amer. Math. Soc ), L. Lewin, Polylogarithms and associated functions, North Holland, New York, , ed., Structural properties of polylogarithms, American Mathematical Society, Providence, RI, 99. Suffolk University, Boston, MA 008 address: knopp00@verizon.net