THE DIRICHLET DIVISOR PROBLEM, BESSEL SERIES EXPANSIONS IN RAMANUJAN S LOST NOTEBOOK, AND RIESZ SUMS. Bruce Berndt

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1 Title and Place THE DIRICHLET DIVISOR PROBLEM, BESSEL SERIES EXPANSIONS IN RAMANUJAN S LOST NOTEBOOK, AND RIESZ SUMS Bruce Berndt University of Illinois at Urbana-Champaign IN CELEBRATION OF THE 80TH BIRTHDAY OF RICHARD ASKEY December 6, / 29

2 Thanks Research Conducted by the Speaker with Sun Kim Ohio State University and Alexandru Zaharescu University of Illinois at Urbana-Champaign 2 / 29

3 Once Again Mystery in the Lost Notebook I have shown you today the highest secret of my own realization. It is supreme and most mysterious indeed. Verse 575, Vivekachudamani of Adi Shankaracharya Sixth Century, A.D. 3 / 29

4 Ramanujan s Passport Photo 4 / 29

5 The Dirichlet Divisor Problem d(n) denotes the number of positive divisors of n. 5 / 29

6 The Dirichlet Divisor Problem d(n) denotes the number of positive divisors of n. Theorem (Dirichlet, 1849) For x > 0, set D(x) := n x 1 d(n) = x(log x + 2γ 1) + + (x), (1) 4 where the prime on the summation sign on the left-hand side indicates that if x is an integer then only 1 2d(x) is counted, γ is Euler s constant, and (x) is the error term. Then, as x, (x) = O( x). (2) 5 / 29

7 The Dirichlet Divisor Problem Figure: The Dirichlet Divisor Problem 6 / 29

8 The Dirichlet Divisor Problem Conjecture For each ɛ > 0, as x, (x) = O(x 1/4+ɛ ). 7 / 29

9 The Dirichlet Divisor Problem Conjecture For each ɛ > 0, as x, (x) = O(x 1/4+ɛ ). Theorem (Hardy, 1916) (x) = Ω + ({x log x} 1/4 log log x). 7 / 29

10 The Dirichlet Divisor Problem Conjecture For each ɛ > 0, as x, (x) = O(x 1/4+ɛ ). Theorem (Hardy, 1916) (x) = Ω + ({x log x} 1/4 log log x). Theorem (Voronoï, 1904; Huxley, 2003) As x, (x) = O(x 1/3 log x), = O(x 131/416+ɛ ) 7 / 29

11 The Dirichlet Divisor Problem Conjecture For each ɛ > 0, as x, (x) = O(x 1/4+ɛ ). Theorem (Hardy, 1916) (x) = Ω + ({x log x} 1/4 log log x). Theorem (Voronoï, 1904; Huxley, 2003) As x, (x) = O(x 1/3 log x), = O(x 131/416+ɛ ) = / 29

12 The Dirichlet Divisor Problem Theorem (Voronoï, 1904) If x > 0, 1 d(n) = x (log x + 2γ 1) ( x ) 1/2 d(n) I1 (4π nx), n n x where I 1 (z) is defined by n=1 I ν (z) := Y ν (z) 2 π K ν(z). (3) 8 / 29

13 A Page From Ramanujan s Lost Notebook 9 / 29

14 Second Bessel Function Series Claim Define F (x) = { [x], if x is not an integer, x 1 2, if x is an integer, (4) where, as customary, [x] is the greatest integer less than or equal to x. 10 / 29

15 Second Bessel Function Series Claim Define F (x) = { [x], if x is not an integer, x 1 2, if x is an integer, (4) where, as customary, [x] is the greatest integer less than or equal to x. I ν (z) := Y ν (z) 2 π K ν(z). (5) 10 / 29

16 Second Bessel Function Series Claim Entry Let F (x) be defined by (4), and let I 1 (x) be defined by (5). For x > 0 and 0 < θ < 1, F n=1 + 1 x 2 ( x ) cos(2πnθ) = 1 x log(2 sin(πθ)) n 4 m=1 n=0 I 1 (4π ) m(n + θ)x + m(n + θ) I 1 (4π ) m(n + 1 θ)x m(n + 1 θ). 11 / 29

17 Second Bessel Function Series Claim As x, 2 ( Y ν (x) = πx sin x πν 2 π 4 π K ν (x) = 2x e x + O ( e x 1 x 3/2 ) ( 1 + O ). x 3/2 ), 12 / 29

18 Second Bessel Function Series Claim As x, 2 ( Y ν (x) = πx sin x πν 2 π 4 π K ν (x) = 2x e x + O ( e x 1 x 3/2 ) ( 1 + O ). x 3/2 ), 1 π 2x 1/4 m ( 3/4 sin 4π ) m(n + θ)x 3 4 π sin (4π ) m(n + 1 θ)x 3 4 π (n + θ) 3/4 + (n + 1 θ) 3/4. 12 / 29

19 Equivalent Theorem with the Order of Summation Reversed Theorem Fix x > 0 and set θ = u + 1 2, where 1 2 < u < 1 2. Recall that F (x) is defined in (4). If the identity below is valid for at least one value of θ, then it exists for all values of θ, and ( x ) ( 1) n F cos(2πnu) 1 + x log(2 cos(πu)) n 4 1 n x = 1 2π 1 n ( + u 2π(n + 1 n=0 M 0 sin lim M 2 + u)x t { M ( 2π(n + 1 sin m=1 ) dt } 2 + u)x m ) 13 / 29

20 Equivalent Theorem with the Order of Summation Reversed Theorem (Continued) + 1 2π n=0 M 0 1 n ( u 2π(n + 1 sin lim M 2 u)x t { M ( 2π(n + 1 sin m=1 ) dt } 2 u)x m ). (6) Moreover, the series on the right-hand side of (6) converges uniformly on compact subintervals of ( 1 2, 1 2 ). 14 / 29

21 Another Interpretation of Ramanujan s Claim Entry Let F (x) be defined by (4), and let I 1 (x) be defined by (5). For x > 0 and 0 < θ < 1, F n=1 + 1 x 2 ( x ) cos(2πnθ) = 1 x log(2 sin(πθ)) n 4 m 1,n 0 I 1 (4π ) m(n + θ)x + m(n + θ) I 1 (4π ) m(n + 1 θ)x m(n + 1 θ). 15 / 29

22 Weighted Divisor Sums If χ denotes any character modulo q, we define d χ (n) := d n χ(d), 16 / 29

23 Weighted Divisor Sums If χ denotes any character modulo q, we define d χ (n) := d n χ(d), ζ(s)l(s, χ) = d χ (n)n s, n=1 16 / 29

24 Weighted Divisor Sums If χ denotes any character modulo q, we define d χ (n) := d n χ(d), ζ(s)l(s, χ) = d χ (n)n s, q 1 n=1 τ(χ) := χ(h)e 2πih/q. h=1 16 / 29

25 Functional Equation for Nonprincipal Even Primitive Characters ( ) π 2s Γ 2 (s)ζ(2s)l(2s, χ) q = τ(χ) ( ) 1 π q 2( 2 s) Γ 2 ( 1 q 2 s) ζ(1 2s)L(1 2s, χ). 17 / 29

26 Hark Back to My Ph.D. Thesis Theorem If χ is a nonprincipal even primitive character modulo q, then dχ (n) = n x q τ(χ) n=1 x d χ (n) n I ( ) 1 4π nx/q x q 1 χ(h) log ( 2 sin(πh/q) ). (7) τ(χ) h=1 18 / 29

27 Summary We have proved Ramanujan s Second Identity with the order of summation reversed. 19 / 29

28 Summary We have proved Ramanujan s Second Identity with the order of summation reversed. We have proved Ramanujan s Second Identity with the product of the summation indices tending to infinity. 19 / 29

29 Summary We have proved Ramanujan s Second Identity with the order of summation reversed. We have proved Ramanujan s Second Identity with the product of the summation indices tending to infinity. We have not proved Ramanujan s Second Identity with the order of summation as written by Ramanujan. 19 / 29

30 Summary We have proved Ramanujan s Second Identity with the order of summation reversed. We have proved Ramanujan s Second Identity with the product of the summation indices tending to infinity. We have not proved Ramanujan s Second Identity with the order of summation as written by Ramanujan. We have proved Ramanujan s First Identity under all three interpretations. 19 / 29

31 Summary We have proved Ramanujan s Second Identity with the order of summation reversed. We have proved Ramanujan s Second Identity with the product of the summation indices tending to infinity. We have not proved Ramanujan s Second Identity with the order of summation as written by Ramanujan. We have proved Ramanujan s First Identity under all three interpretations. Road Blocks: The plus sign between the two Bessel functions; singularities at / 29

32 Summary We have proved Ramanujan s Second Identity with the order of summation reversed. We have proved Ramanujan s Second Identity with the product of the summation indices tending to infinity. We have not proved Ramanujan s Second Identity with the order of summation as written by Ramanujan. We have proved Ramanujan s First Identity under all three interpretations. Road Blocks: The plus sign between the two Bessel functions; singularities at 0. The proofs under different interpretations are completely different. 19 / 29

33 Summary We have proved Ramanujan s Second Identity with the order of summation reversed. We have proved Ramanujan s Second Identity with the product of the summation indices tending to infinity. We have not proved Ramanujan s Second Identity with the order of summation as written by Ramanujan. We have proved Ramanujan s First Identity under all three interpretations. Road Blocks: The plus sign between the two Bessel functions; singularities at 0. The proofs under different interpretations are completely different. New methods of estimating trigonometric sums are introduced. 19 / 29

34 Riesz Sums a(n)(x n) a n x 20 / 29

35 Analogue of a Theorem of Dixon and Ferrar Theorem Let a denote a positive integer, and let χ be an even primitive nonprincipal character of modulus q. Set for x > 0, Then, for a 2, D χ (a; x) := 1 Γ(a + 1) d χ (n)(x n) a. n x L(1, χ)x a D χ (a 1; x) = Γ(1 + a) + τ(χ)qa 1 (2π) a 1 ( nx { Y a 4π q n=1 ) + 2 π cos(πa)k a ( x d χ (n) nq ( 4π nx q ) a/2 )} + S a, 21 / 29

36 Analogue of a Theorem of Dixon and Ferrar Theorem (Continued) where S a is the sum of the residues at the poles 2m 1, 0 m [ 1 2a] 1, of Γ(s)ζ(s)L(s, χ) x s+a 1. Γ(s + a) 22 / 29

37 The Riesz Sum Generalization of Ramanujan s Entry 5 Theorem Let x > 0, 0 < θ < 1, and a be a positive integer. Then, 1 (x n) a 1 cos(2πrθ) = x a 1 (a 1)! 4(a 1)! x a a! log ( 2 sin(πθ) ) n x r n + x α/2 I a (4π ) m(n + θ)x I a (4π ) m(n + 1 θ)x 2(2π) a 1 ( ) a/2 + ( ) a/2 m=1 n=0 m(n + θ) m(n + 1 θ) ( ) [a/2] ( 1) k ζ(1 2k) ζ(2k, θ) + ζ(2k, 1 θ) x a 2k (a 2k)!(2π) 2k, (8) k=1 where I ν (x) is defined in (3), and ζ(s, α) denotes the Hurwitz zeta function. 23 / 29

38 An Analogue of Ramanujan s Entry 5 Theorem Let I 1 (x) be defined by (5). If 0 < θ, σ < 1 and x > 0, then 1 x cos(2πnθ) cos(2πmσ) = nm x n,m 0 { I 1 (4π (n + θ)(m + σ)x) + I 1(4π (n + 1 θ)(m + σ)x) (n + θ)(m + σ) (n + 1 θ)(m + σ) + I 1(4π (n + θ)(m + 1 σ)x) (n + θ)(m + 1 σ) + I 1(4π } (n + 1 θ)(m + 1 σ)x). (n + 1 θ)(m + 1 σ) 24 / 29

39 An Analogue of Ramanujan s Entry 5 Define, for Dirichlet characters χ 1 modulo p and χ 2 modulo q, d χ1,χ 2 (n) = χ 1 (d)χ 2 (n/d). d n 25 / 29

40 An Analogue of Ramanujan s Entry 5 Define, for Dirichlet characters χ 1 modulo p and χ 2 modulo q, d χ1,χ 2 (n) = χ 1 (d)χ 2 (n/d). d n L(2s, χ 1 )L(2s, χ 2 ) = n=1 χ 1 (n) n 2s m=1 χ 2 (m) m 2s = n=1 d χ1,χ 2 (n) n 2s 25 / 29

41 An Analogue of Ramanujan s Entry 5 Define, for Dirichlet characters χ 1 modulo p and χ 2 modulo q, d χ1,χ 2 (n) = χ 1 (d)χ 2 (n/d). d n L(2s, χ 1 )L(2s, χ 2 ) = n=1 χ 1 (n) n 2s m=1 χ 2 (m) m 2s = The Functional Equation n=1 d χ1,χ 2 (n) n 2s (π 2 /(pq)) s Γ 2 (s)l(2s, χ 1 )L(2s, χ 2 ) = τ(χ 1)τ(χ 2 ) pq (π 2 /(pq)) ( 1 2 s) Γ 2 ( 1 2 s)l(1 2s, χ 1)L(1 2s, χ 2 ) 25 / 29

42 An Analogue of Ramanujan s Entry 5 Define, for Dirichlet characters χ 1 modulo p and χ 2 modulo q, d χ1,χ 2 (n) = χ 1 (d)χ 2 (n/d). d n L(2s, χ 1 )L(2s, χ 2 ) = n=1 χ 1 (n) n 2s m=1 χ 2 (m) m 2s = The Functional Equation n=1 d χ1,χ 2 (n) n 2s (π 2 /(pq)) s Γ 2 (s)l(2s, χ 1 )L(2s, χ 2 ) = τ(χ 1)τ(χ 2 ) pq (π 2 /(pq)) ( 1 2 s) Γ 2 ( 1 2 s)l(1 2s, χ 1)L(1 2s, χ 2 ) dχ1,χ 2 (n) = τ(χ 1)τ(χ 2 ) pq n x n=1 ( x ) 1 2 nx ) d χ1,χ 2 (n) I 1 (4π n pq 25 / 29

43 Hark back to My Ph.D. Thesis I had trouble estimating a crucial integral. 26 / 29

44 Hark back to My Ph.D. Thesis I had trouble estimating a crucial integral. It was suggested that I see Professor Richard Askey, 26 / 29

45 Hark back to My Ph.D. Thesis I had trouble estimating a crucial integral. It was suggested that I see Professor Richard Askey, who suggested that I see Professor Steve Wainger, 26 / 29

46 Hark back to My Ph.D. Thesis I had trouble estimating a crucial integral. It was suggested that I see Professor Richard Askey, who suggested that I see Professor Steve Wainger, who suggested that I learn the method of steepest descent, and so I was able to complete my dissertation. 26 / 29

47 Hark back to My Ph.D. Thesis I had trouble estimating a crucial integral. It was suggested that I see Professor Richard Askey, who suggested that I see Professor Steve Wainger, who suggested that I learn the method of steepest descent, and so I was able to complete my dissertation. Identities involving the coefficients of a class of Dirichlet series. I, Trans. Amer. Math. Soc. 137 (1969), / 29

48 Hark back to My Ph.D. Thesis I had trouble estimating a crucial integral. It was suggested that I see Professor Richard Askey, who suggested that I see Professor Steve Wainger, who suggested that I learn the method of steepest descent, and so I was able to complete my dissertation. Identities involving the coefficients of a class of Dirichlet series. I, Trans. Amer. Math. Soc. 137 (1969), T. J. Kaczynski (The Unabomber), Boundary functions for bounded harmonic functions, Trans. Amer. Math. Soc. 137 (1969), / 29

49 Photograph, Shanghai, July 31, / 29

50 Photograph, Shanghai, July 31, / 29

51 Happy Birthday THE DIRICHLET DIVISOR PROBLEM FOR DICK ASKEY 29 / 29

52 Happy Birthday THE DIRICHLET DIVISOR PROBLEM FOR DICK ASKEY Happy 80th Birthday, Dick 29 / 29

53 Happy Birthday THE DIRICHLET DIVISOR PROBLEM FOR DICK ASKEY Happy 80th Birthday, Dick and Many more 29 / 29