Yet another failed attempt to prove the Irrationality of ζ(5)

Transcription

1 Yet another failed attempt to prove the Irrationality of ζ5 Abstract Introduction The irrationality of the values of the Riemann zeta function at even integers has been known since Euler. However, for odd integers very little is known. A breakthrough came in 978 when R. Apery presented his proof that ζ3 is irrational []. Shortly thereafter F. Beukers presented a much more accessible proof of this result [4]. In the past three decades many papers have appeared studying the irrationality of ζn + using Beukers type integrals and the closely related Sorokin type integrals. In many of them linear combinations of the values ζn for n > over Q are examined. These yield to lower bounds of the dimensions of the Q vector spaces spanned by values of the zeta function. T. Rivoal and K. Ball showed that infinitely many of the values ζn+ are irrational [0, 3], and W. Zudilin later showed that at least one of the numbers ζ5, ζ7, ζ9, ζ is irrational [4]. Works by T. Rivoal, S. Fischler, Yu. Nesterenko, C. Viola, and W. Zudilin improved these techniques and results see, for example, [5, 7,, 9, 3]. But to date, the irrationality for a specific value of ζn+ other than ζ3 has not been shown. In this note we explore ζ5 via two different approaches using Beukers type integrals. While both approaches ultimately fail to produce the desired result, the nature of this failure comes at the opposite end of the arguments. This may give some information about how to approach this problem. Our approaches are based on the following easily proved identity: logx x x 3 x 4 ζ5 dx dx dx 3 dx 4 0, logx x 3 dx dx 0, 4 x x x 3 x 4 6 x x In section of this note we will use the double integral to show that a positive linear expression involving only ζ5 and integers can be expressed by a Beukers type integral. We believe that this approach is new and can offer some valuable insights. Unfortunately, this integral does not give the desired upper bound which allows one to show that the expression will go to zero as N increases. This

2 approach was inspired by J. Sondow s work on irrationality conditions for Euler s constant [], and D. Huylebrouck s paper [8]. In section 3 we will use an approach based on the quadruple integral. While this approach yields excellent upper bounds, one only gets them for a positive linear combination of ζ3, ζ4, and ζ5 over Z. Take, the double integral approach The main result here is Proposition For every N N there exist integers A N and B N such that logx x 3 p N x p N x 0 < A N ζ5+b N dx dx d 5 N 0, x x, where d N lcm,,...,n, and p N x denotes the Legendre polynomial of degree N over [0,]. We split the proof of this result into two Lemmas: Lemma Let r,s N. Let d r lcm,,...r. We have: where d r q r Z. logx x 3 x r xs dx dx 0, x x 4 ζ5 r k k 5 q r Q if r s if r > s 3 Proof: The proof follows the proof of Lemma in [] We start by noting that logx x 3 d3 dǫ 3x x ǫ. ǫ0 Replacing the logarithm in the integral by the expression on the right hand side of this last equation and expanding the geometric series one has logx x 3 x r xs 0, x x dx dx d3 dǫ 3 d3 dǫ 3 k0 k0 x r+ǫ x s+ǫ dx dx 0, ǫ0 k ++r+ǫk ++s+ǫ If r > s, the last sum will collapse and yield { d3 dǫ 3 r s s++ǫ + + r +ǫ} ǫ0 ǫ0

3 Evaluating the third derivative will give logx x 3 x r xs dx dx 6 0, x x r s When r s, we take the third derivative and evaluate at ǫ 0 to get logx x 3 x r xr dx dx 4 0, x x and the proof of the Lemma is complete. To continue we show k0 { s } r 4 k ++r 5 4 kr+ k 5 Lemma There exist integers A n and B N such that logx x 3 p N x p N x 0 < dx dx A N ζ5+b N d 5 N 0, x x Proof: Since the Legendre polynomials p N x,p N x have integral coefficients, the equality follows from the previous lemma. To show the positivity one first eliminates the logarithms by writing logx x 3 p N x p N x logx x 3 x x p N x p N x dx dx 0, x x 0, x x 3 dx dx and using the identity three times to get logx x x x 0 dz x x z logx x 3 p N x p N x dx dx 0, x x x x p N x p N x 0, 5 x x z x x z x x z 3 dx dx dz dz dz 3. The critical part of Beukers proof is the next sequence of integration by parts, followed by substitution followed by a second integration by parts. The more complicated denominator and extra factor in the numerator complicate this process. An easy induction argument shows that n x n x x z x x z x x z 3 n n!x n z izj zk 3 x x z i+ x x z j+ x x z 3 k+ i+j+kn n n!x n Q n. 3

4 It follows that n x n x x x x z x x z x x z 3 x x n n!x n Q n 4n x x x n n!x n Q n +4 nn x n n!x n Q n n n!x n x x Q n +4 x x Q n +Q n. Observe that all terms inside the prentices are positive and it is easily seen that the sums defining Q n have n+n+/ terms and so the number of terms grows like a second degree polynomial in n. Integrating by parts in our original integral N times will produce 0, 5 x x Q N +4 x x Q N +Q N x N x N x! N P N x dx dx dz dz dz 3 We continue by making the substitutions w i z i x x z i for i,,3. This has the effect that terms of the form z i zj zk 3 x x z i+ x x z j+ x x z 3 k+ w i w j w 3 k x x w x x w x x w 3 So the new integrand can be written as x x R N +4 x x R N +R N x N P N x, x x w x x w x x w 3 where R N i+j+kn w i w j w 3 k Except for the R N part of this the three terms are similar, and in one case equal to the original integrand. The R N are unaffected by the integration by parts in x. Integration by parts N times in x will yield the following integrand I N R N x x Q N +4 x x Q N +Q N +R N x x Q N +Q N +R N Q N x x x x N. 4

5 This integrant is clearly positive, which conludes the proof of the lemma and the proposition. To prove the irrationality of ζ5, one must now show that this integral converges to zero at a rate faster than the exponential growth of d 5 N. From the Prime Number Theorem we know that d N < e +ǫn, which yields a growth for d 5 N which is faster than e5n. To look at the growth of our integrand we write I N J N x x w x x w x x w 3 x x x x N J N is sum of q N terms, where q N is a polynomial of degree four. The terms of J N are of the form w i w j w 3 k w s wt,wu 3 x x w s x x w t x x w 3 u, where i +j +k {N,N,N} and s+t+u {N,N,N}. Since it adds only a fixed factor we may assume that i + j + k N and s + t + u N. Next, observe that all combinations of the sums i + j + k and s + t + u actually appear in J N so that J N is invariant under any group action of S 3 which permutes the values of w, w, and w 3. So if I N attaines a maximal value at some particular point x 0,x0,w0,w0,w0 3 it will attain that same value at x0,x0,w0 σ,w0 σ,w0 σ3 for any permutation σ S 3. To continue, let us supposethat the maximal value is attained along the diagonal w w w 3, for if it is not the actual maximal value is certainly at least as large as the maximum along the diagonal. Replacing the individual values of w i in J N by a common w, the terms of J N all become w N w N x x w N which together with thefactors x N x N x N x N combines to the original integrand in Beukers proof of the irrationality of ζ3. Beukers provided an upper bound for this integrand in the form of C+ 4N. Combining this with the fact that the actual maximum may excede the value on the diagonal we get Lemma 3 If there is a real number ρ such that 0, logx x 3 p N x p N x x x dx dx Kq N ρ N, for some constant K and the polynomial q N obtained in the derivation above, then ρ + 4 5

6 Since + 4 < e 5, this approach cannot possibly yield a proof for the irrationality of ζ5. However, it came as a real surprise to the author that the critical steps of the proof of Lemma, namely the sequence of integration by parts, followed by substitution, followed by another integration by parts, still work without returning expressions which are not managable. 3 Take, the quadruple integral approach Another way of trying is to use the quadruple integral representation mentioned in the introduction. This attempt will fail as well, however, the failure will be at a different level of the proof. It will reproduce the known result: Proposition There are integer sequences A N, B N, C N, D N such that 0 < A N ζ5+b N ζ4+c N ζ3+d N < K3 N, for some constant K. Moreover, if d N lcm,...,n then d 5 N A N, d 4 N B N, D 3 N C N. As in the previous section we divide the proof of this proposition into several lemmas. Moreover, as we have a quadruple integral to play with we will not use the notations x i,i,...,4, but rather x, y, u, v, as the x,y variables are handled very differently than the u,v ones. Lemma 4 Let s, s, s 3, and s 4 be nonnegative integers and r max{s,s,s 3,s 4 }. Moreover, let d r lcm,...,r. Then where logxyuv xyuv xs y s u s 3 v s 4 dxdydudv 4 4 ζ5 r k k 5 q Q p ζ3+q p 3 ζ3+q 3 p 4 ζ4+p 4 ζ3+q 4 if s s s 3 s 4 r if the four exponents are pairwise distinct if there are three distinct exponents if there are two distinct exponents appearing twice each if there are two distinct exponents, with one appearing three times, 5 d 5 rq i Z 6 6

7 Proof: Following the lines of the previous section we write d 3 r p,d r p 3,d r p 4 Z 7 d r p 4 Z 8 logxyuv d dǫ xyuvǫ ǫ0 and expand the integrand into a power series. After integration we can look at the different cases. Case : s s s 3 s 4. In this case the terms of the series are given by k++s +ǫ 4 and differentiation and evaluating at ǫ 0 will yield the expression involing ζ5. Case : Four pairwise distinct exponents. In this case we have the typical term k++s +ǫk ++s +ǫk ++s 3 +ǫk ++s 4 +ǫ s s s 3 s 4 k ++s +ǫ k ++s +ǫ k++s 3 +ǫ k++s 4 +ǫ s s s 3 s 4 k ++s +ǫk ++s 3 +ǫ k++s +ǫk ++s 3 +ǫ + k ++s +ǫk ++s 4 +ǫ k ++s +ǫk ++s 4 +ǫ Partial fraction decomposition can be done again resulting in four collapsing sums of the form s s s 4 s 3 s i s j k ++s j +ǫ. k ++s i +ǫ k0 Since these sums are finite, differentiation and evaluating the result at ǫ 0 will result in a rational number with a denominator with the stated properties. Case 3: If there are three distinct exponents the terms of the sum are k++s +ǫ k++s +ǫk ++s 3 +ǫ s s s 3 s k ++s +ǫ k ++s +ǫ k++s +ǫ k++s 3 +ǫ s s s 3 s k ++s +ǫk ++s +ǫ k++s +ǫk ++s 3 +ǫ + k ++s +ǫk ++s 3 +ǫ k ++s +ǫk ++s 3 +ǫ All but the first term can be further decomposed yielding collapsing sums as in Case above. The first term will yield a rational multiple of ζ3 with a denominator with the desired properties. 7

8 Case 4: Two distinct exponents appearing twice each. Of the terms k ++s +ǫ k ++s +ǫ s s k ++s +ǫ k ++s +ǫ s s k ++s +ǫ k ++s +ǫk ++s +ǫ + k++s +ǫ the middle one will yield a collapsing sum after partial fraction decomposition, and the quadratic terms will yield rational multiples of ζ3. Case 5: One exponent appears three times. We have k ++s +ǫ 3 k ++s +ǫ s s k ++s +ǫ k ++s +ǫ k++s +ǫ s s k ++s +ǫ 3 k++s +ǫk ++s +ǫ The first term will give rise to a rational multiple of ζ4, and the second one will yield a ζ3 term and a rational number after partial fraction decomposition, which completes the proof of the lemma. Lemma 5 There exists integers A N, B N, C N, and D N such that for some positive constant K 0 < A N ζ5+b N ζ4+c N ζ3+d N d 5 logxyuv u N v N p N xp N y N dxdydudv 0, 4 xyuv < K3 N, Proof: The proof will be rather sketchy as it involves much of the same steps as the proof in the previous section. After eliminating the logarithm using a similar integral as in the previous section and going through the sequence of integration by parts in x followed by substitution, followed by integration by parts in y the integral in the Lemma becomes x xy yu uv v N dxdydudvdw xyuvw xyuvw. 0, 5 Thus the expression is positive. To estimate the integrant observe that x xy y xyuvw x xy y xyw < + 4 8

9 by Beukers argument. Furthermore, u uv v 6. Therefore the integral can be esimated as follows: x xy yu uv v Observe that 0, 5 which yields the desired result. xyuvw < 6+ 4 N 6+ 4 Nζ5 N dxdydudvdw xyuvw dxdydudvdw 0, 5 xyuvw 6+ 4 > 543 > 3 73 > > 3d 5 N 4 Lessons learned from the repeated exercise in futility While both approaches ultimately fail there are some valuable lessons from each one. First, it was positive surprise that the sequence of operations involving the transformations of integrals to show positivity is very robust and can handle more complex integrands without deteriorating into an incredible mess. This leaves hope for different scenarios. Secondly, it appears that the quality of the estimate of the integral improves rapidly if the number of variables increases. However, the increase of numbers of variables needs to be paid for by the appearance of undesirable terms involving other values of ζ. On could try to use an approach based on the integral 0,3 logx x x 3 dx dx dx 3 x x x 3 This would be a compromise of the two approaches presented above. However, since there would be terms in which two of the exponents are equal one will certainly pick up a term that involves ζ3. And in the estimate one would get that the integral is bounded by K 4+sqrt 4 N which doesn t outway the factor d 5 N. One glimmer of hope is that the coefficients that appear in the linear combinations of ζ values in Proposition can be explicitely computed and have resemblences with the Apery numbers that appear in the argument for ζ3. Using several such linear combinations one may be able to eliminate the undesireable values of ζ while preserving both the positivity and a rapidly decreasing upper bound. But that looks like an incredible mess to begin with. 9

10 References [] Andrews, G. E., Askey, R., Roy, R.,Special Functions, Encyclopedia of Mathematics and its Applications 7, Cambridge University Press, Cambridge, UK, 999. [] Apéry, R., Irrationalité de ζ et ζ3, Asterisque 6, 3, 979. [3] Ball, K., Rivoal, T., Irrationalite d une infinité de valeurs de la fonction zeta aux entiers impairs, Invent. math. 46, 93-07, 00. [4] Beukers, F., A Note on the Irrationality of ζ and ζ3, Bull. London Math. Soc., 68 7, 979. [5] Fischler, F., Restricted rational approximation and Apéry-type constructions, Indag. Mathem., N.S. 0, 0 5, 009. [6] Fischler, F., Rivoal, T., Irrationality exponent and rational approximations with prescribed growth, Proc. Amer. Math. Soc. 383, , 00. [7] Fischler, F., Zudilin, W., A refinement of Nesterenko s linear independece criterion with applications to zeta values, Math. Ann. 347, , 00. [8] Huylebrouck, D., Similarities in Irrationality Proofs for π, ln, ζ, and ζ3, Amer. Math. Monthly 8, 3, 00. [9] Nesterenko, Yu., A few remarks on ζ3, Math. Notes 59, , 996. [0] Rivoal, T., La fonction zeta de Riemann prende une infinité de valeurs irrationelles aux entiers impairs entiers, Comptes Rendus de l Academie des Sciences. Serie Mathematique 33, 67 70, 000. [] Rivoal, T., Linear forms in zeta values arising from certain Sorokin-type integrals, J. of Math. Sciences 805,64 649, 0. [] Sondow, J., Criteria for the irrationality of Euler s constant, Proc. Amer. Math. Soc. 3, , 003. [3] Viola, C., On the equivalence of Beukers-type and Sorokin-type multiple integrals, J. of Math. Sci. 805, , 0. 0

11 [4] Zudilin, W., One of the numbers ζ5, ζ7, ζ9, ζ is irrational, Rus. Math. Sur. 564, , 00.