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Sums of Generalized Harmonic Series for Kids from Five to Fifteen

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Sums of Generalized Harmonic Series for Kids from Five to Fifteen GENERAL ⎜ ARTICLE Sums of Generalized Harmonic Series For Kids from Five to Fifteen Zurab Silagadze We examine the remarkable connection, first dis- covered by Beukers, Kolk and Calabi, between ζ(2n), the value of the Riemann Zeta-function at an even positive integer, and the volume of some 2n-dimensional polytope. It can be shown that this volume is equal to the trace of a compact Zurab Silagadze gradu- self-adjoint operator. We provide an explicit ex- ated from the Tbilisi State pression for the kernel of this operator in terms University in 1979. He got of Euler polynomials. This explicit expression his PhD from Novosibirsk in theoretical and makes it easy to calculate the volume of the poly- mathematical physics. tope and hence ζ(2n). In the case of odd positive Currently, he works as integers, we rediscover an integral representa- senior researcher at tion for ζ(2n +1), obtained by a different method Budker Institute of by Cvijovi´c and Klinowski. Finally, we indicate Nuclear Physics and as assistant professor at that the origin of the miraculous Beukers–Kolk– Novosibirsk State Calabi change of variables in the multidimen- University. sional integral, which is at the heart of this circle of ideas, can be traced to the amoeba associated with the certain Laurent polynomial. The article is dedicated to the memory of Vladimir Arnold (1937–2010). 1. Introduction In a nice little book [1], Vladimir Arnold has collected 77 mathematical problems for kids from 5 to 15 to stimulate critical thinking in them. Problem 51 in this book asks the reader to calculate the sum of the inverse squares and prove Euler’s celebrated formula ∞ 1 π2 2 = . (1) Keywords n 6 n=1 Riemann zeta function, integral representation, Basel problem. Well, there are many waystodothis(see, for example, 822 RESONANCE ⎜ September 2015 GENERAL ⎜ ARTICLE [2, 3, 4, 5, 6, 7, 8, 9, 10] and references therein), some maybe even accessible for kids under fifteen1. However, 1 Editor’s note: This perhaps ap- in this note we concentrate on the approach of Beuk- plies to Russian kids! ers, Kolk and Calabi [11], further elaborated by Elkies in [12]. This approach incorporates pleasant features which all the kids (and even some adults) adore: sim- plicity, magic and the depth that allows one to go be- yond the particular case (1). The simplicity, however, is not everywhere explicit in [11] and [12], while the magic longs for explanation after the first admiration fades away. Below we will try to enhance the simplicity of the approach and somewhat uncover the secret of the magic. The article is organized as follows. In the first two sec- tions, we reconsider the evaluation of ζ(2) and ζ(3) so that technical details of the general case do not obscure the simple underlying ideas. Then, we elaborate the general case and give the main result of this work, the formula for the kernel which allows us to simplify consid- erably the evaluation of ζ(2n) from [11, 12] and re-derive Cvijovi´c and Klinowski’s integral representation [13] for ζ(2n + 1). Finally, we ponder over the mysterious re- lations between the sums of generalized harmonic series and amoebas, first indicated by Passare in [10]. This relation enables us to uncover somewhat the origin of the Beukers–Kolk–Calabi’s highly non-trivial change of variables. The approach of Evaluation of ζ(2) Beukers, Kolk and Calabi has the Recall the definition of the Riemann Zeta function: features of simplicity, ∞ 1 magic and depth. ζ(s)= . (2) But, the simplicity is ns n=1 not explicit and the The sum (1) is just ζ(2) which we will now evaluate magic longs for following the method of Beukers, Kolk and Calabi [11]. explanation after the first admiration fades away. RESONANCE ⎜ September 2015 823 GENERAL ⎜ ARTICLE This change of Our starting point will be the dilogarithm function variables ∞ xn transforms the unit Li (x)= . (3) 2 n2 square to an n=1 isosceles triangle. Clearly, Li2(0) = 0 and Li2(1) = ζ(2). Differentiating (3), we get ∞ d xn x Li (x)= = − ln (1 − x), dx 2 n n=1 and, therefore, 1 ln (1 − x) dx dy ζ(2) = Li2(1) = − dx = , (4) x 1 − xy 0 ✷ where ✷ = {(x, y):0≤ x ≤ 1, 0 ≤ y ≤ 1} is the unit square. Let us note dx dy dx dy dx dy + =2 , (5) 1 − xy 1+xy 1 − x2y2 ✷ ✷ ✷ and dx dy dx dy 1 dx dy − = , (6) 1 − xy 1+xy 2 1 − xy ✷ ✷ ✷ where the last equation follows from 2xy 1 d(x2)d(y2) 1 dx dy dx dy = = . 1 − x2y2 2 1 − x2y2 2 1 − xy ✷ ✷ ✷ It follows from equations (5) and (6) that 4 dx dy ζ(2) = . (7) 3 1 − x2y2 ✷ Now let us make the magic Beukers–Kolk–Calabi change of variables in this two-dimensional integral [11] sin u sin v x = ,y= , (8) cos v cos u 824 RESONANCE ⎜ September 2015 GENERAL ⎜ ARTICLE with Jacobian determinant cos u sin v sin u cos v cos2 u 2 2 ∂(x, y) sin u sin v 2 2 = =1− 2 2 =1−x y . ∂(u, v) sin u sin v cos v cos v cos u 2 cos v cos u Then miraculously Editor's note: For a change of variables, the integral gets 4 4 scaled by the Jacobian determi- ζ(2) = du dv = Area(∆), (9) nant. 3 3 ∆ where ∆ is the image of the unit square ✷ under the transformation (x, y) → (u, v). It is easy to show that ∆ is the isosceles right triangle ∆ = {(u, v):u ≥ 0,v ≥ 0,u+ v ≤ π/2} and, therefore, 4 1 π 2 π2 ζ(2) = = . (10) 3 2 2 6 “Beautiful – even more so, as the same method of proof extends to the computation of ζ(2k)intermsofa2k- dimensional integral, for all k ≥ 1” [14]. However, before considering the general case, we check whether the trick works for ζ(3). 3. Evaluation of ζ(3) Inthecaseofζ(3), we begin with trilogarithm ∞ xn Li (x)= , (11) 3 n3 n=1 and using x d ln (1 − y) x Li3(x)=Li2(x)=− dy, dx y 0 we get 1 x dx ln (1 − y) ζ(3) = Li3(1) = − dy. (12) x y 0 0 RESONANCE ⎜ September 2015 825 GENERAL ⎜ ARTICLE A change of variables But of the unit cube x 1 1 ln (1 − y) ln (1 − xz) similar to that of the − dy = − dz square unfortunately x y xz 0 0 leads to a 1 1 complicated integral. dy = dz , A hyperbolic version 1 − xyz gives an interesting 0 0 integral which too is and finally not possible to dx dy dz evaluate; hence ζ(3) = Li3(1) = , (13) 1 − xyz zeta(3) is more ✷3 complicated than ✷ zeta(2). where 3 = {(x,y,z):0≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1} is the unit cube. By a similar trick as before, we can transform (13) into the integral 8 dx dy dz ζ(3) = , (14) 7 1 − x2y2z2 ✷3 and here the analogy with the previous case ends, un- fortunately, because the generalization of the Beukers– Kolk–Calabi change of variables does not lead in this case to a simple integral. However, it is interesting to note that the hyperbolic version of this change of vari- ables sinh u sinh v sinh w x = ,y= ,z= (15) cosh v cosh w cosh u does indeed produce an interesting result 8 8 ζ(3) = du dv dw = Vol(U3), (16) 7 7 U3 where U3 is a complicated 3-dimensional shape defined by the inequalities u ≥ 0,v≥ 0,w≥ 0, sinh u ≤ cosh v, sinh v ≤ cosh w, sinh w ≤ cosh u. 826 RESONANCE ⎜ September 2015 GENERAL ⎜ ARTICLE Unfortunately, there is no obvious simple way to calcu- late the volume of U3. However, there is a second way to convert the integral (12) for ζ(3) in which the Beukers–Kolk–Calabi change of variables still plays a helpful role. We begin with the identity 1 x dx ln (1 − y) ln (1 − y) ζ(3) = − dy = − dx dy, x y xy 0 0 D (17) where the domain of the 2-dimensional integration is the triangle D = {(x, y):x ≥ 0,y ≥ 0,y ≤ x}. Interchang- ing the order of integration in (17), we get 1 1 ln (1 − y) dx ζ(3) = − dy , y x 0 y which can be transformed further as follows 1 1 1 ln (1 − y)lny dx ζ(3) = dy = − ln y dy y 1 − xy 0 0 0 ln y = − dx dy, 1 − xy ✷ or in a more symmetrical form 1 ln (xy) ζ(3) = − dx dy. (18) 2 1 − xy ✷ Note that 2xy ln (xy) 1 ln (x2y2) dx dy = d(x2)d(y2) 1 − x2y2 4 1 − x2y2 ✷ ✷ 1 ln (xy) = dx dy. 4 1 − xy ✷ RESONANCE ⎜ September 2015 827 GENERAL ⎜ ARTICLE Therefore, we can modify (5) and (6) accordingly and using them transform (18) into 4 ln (xy) ζ(3) = − dx dy. (19) 7 1 − x2y2 ✷ At this point we can use the Beukers–Kolk–Calabi change of variables (8) in (19) and as a result we get 4 ζ(3) = − ln (tan u tan v)du dv 7 ∆ 8 = − ln (tan u)du dv. (20) 7 ∆ But this equation indicates that π/2 π/2−u 8 ζ(3) = − du ln (tan u) dv 7 0 0 π/2 8 π = − − u ln (tan u)du, 7 2 0 π which after the substitution x = 2 − u becomes π/2 π/2 8 8 ζ(3) = − x ln (cot x)dx = x ln (tan x)dx.
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