A positivity conjecture related to the Riemann zeta function Thomas - PowerPoint PPT Presentation

A positivity conjecture related to the Riemann zeta function Thomas Ransford Universit´ e Laval, Quebec City, Canada Conference in honor of John Garnett and Don Marshall University of Washington, Seattle, August 2019.

Collaborators Hugues Bellemare Yves Langlois Article to appear in the American Mathematical Monthly

The functions f k For each integer k ≥ 2, define f k : (0 , 1) → R by � 1 f k ( x ) := 1 � 1 � � − . k x kx

The functions f k For each integer k ≥ 2, define f k : (0 , 1) → R by � 1 f k ( x ) := 1 � 1 � � − . k x kx Graph of f 5 ( x )

A least-squares problem For g : (0 , 1) → C , we write �� 1 � 1 / 2 | g ( x ) | 2 dx � g � := . 0 Set n � � � d n := min � 1 − λ k f k � . � � λ 2 ,...,λ n k =2 Does d n → 0 as n → ∞ ?

A least-squares problem For g : (0 , 1) → C , we write �� 1 � 1 / 2 | g ( x ) | 2 dx � g � := . 0 Set n � � � d n := min � 1 − λ k f k � . � � λ 2 ,...,λ n k =2 Does d n → 0 as n → ∞ ? Theorem (Nyman 1950, B´ aez-Duarte 2003) d n → 0 ⇐ ⇒ RH

The Riemann hypothesis Euler (1748): For s > 1, 1 1 � � 1 − p − s = n s . p n ≥ 1

The Riemann hypothesis Euler (1748): For s > 1, 1 1 � � 1 − p − s = n s . p n ≥ 1 Riemann (1859): For s complex, Re s > 1, � 1 � 1 1 � 1 s � 1 � x s − 1 dx = � x s − 1 dx . � ζ ( s ) := n s = s s − 1 − s x x 0 0 n ≥ 1

The Riemann hypothesis Euler (1748): For s > 1, 1 1 � � 1 − p − s = n s . p n ≥ 1 Riemann (1859): For s complex, Re s > 1, � 1 � 1 1 � 1 s � 1 � x s − 1 dx = � x s − 1 dx . � ζ ( s ) := n s = s s − 1 − s x x 0 0 n ≥ 1 Riemann hypothesis ζ ( s ) � = 0 for all s with Re s > 1 / 2.

Proof that d n → 0 ⇒ RH A simple calculation shows that, if Re s > 0, then � 1 f k ( x ) x s − 1 dx = ζ ( s ) � 1 k − 1 � . s k s 0

Proof that d n → 0 ⇒ RH A simple calculation shows that, if Re s > 0, then � 1 f k ( x ) x s − 1 dx = ζ ( s ) � 1 k − 1 � . s k s 0 Suppose that ζ ( s 0 ) = 0, where Re s 0 > 1 / 2. Then � 1 f k ( x ) x s 0 − 1 dx = 0 ( k = 2 , 3 , . . . ) . 0

Proof that d n → 0 ⇒ RH A simple calculation shows that, if Re s > 0, then � 1 f k ( x ) x s − 1 dx = ζ ( s ) � 1 k − 1 � . s k s 0 Suppose that ζ ( s 0 ) = 0, where Re s 0 > 1 / 2. Then � 1 f k ( x ) x s 0 − 1 dx = 0 ( k = 2 , 3 , . . . ) . 0 Thus, for all λ 2 , . . . , λ n , � 1 � 1 n x s 0 − 1 dx = 1 � � x s 0 − 1 dx = � 1 − λ k f k ( x ) . s 0 0 0 k =2

Proof that d n → 0 ⇒ RH A simple calculation shows that, if Re s > 0, then � 1 f k ( x ) x s − 1 dx = ζ ( s ) � 1 k − 1 � . s k s 0 Suppose that ζ ( s 0 ) = 0, where Re s 0 > 1 / 2. Then � 1 f k ( x ) x s 0 − 1 dx = 0 ( k = 2 , 3 , . . . ) . 0 Thus, for all λ 2 , . . . , λ n , � 1 � 1 n x s 0 − 1 dx = 1 � � x s 0 − 1 dx = � 1 − λ k f k ( x ) . s 0 0 0 k =2 But also, by Cauchy–Schwarz, � 1 n n � � � x s 0 − 1 dx � � � � � � � x s 0 − 1 � . 1 − λ k f k ( x ) � ≤ � 1 − λ k f k � � � � � 0 k =2 k =2

Proof that d n → 0 ⇒ RH A simple calculation shows that, if Re s > 0, then � 1 f k ( x ) x s − 1 dx = ζ ( s ) � 1 k − 1 � . s k s 0 Suppose that ζ ( s 0 ) = 0, where Re s 0 > 1 / 2. Then � 1 f k ( x ) x s 0 − 1 dx = 0 ( k = 2 , 3 , . . . ) . 0 Thus, for all λ 2 , . . . , λ n , � 1 � 1 n x s 0 − 1 dx = 1 � � x s 0 − 1 dx = � 1 − λ k f k ( x ) . s 0 0 0 k =2 But also, by Cauchy–Schwarz, � 1 n n � � � x s 0 − 1 dx � � � � � � � x s 0 − 1 � . 1 − λ k f k ( x ) � ≤ � 1 − λ k f k � � � � � 0 k =2 k =2 Hence d n ≥ 1 / | s 0 |� x s 0 − 1 � for all n . In particular d n �→ 0.

Graph of d n against n (B´ aez-Duarte et al, 2000)

Graph of d − 2 against log n (Ba´ ez-Duarte et al, 2000) n

Behavior of the sequence ( d n ) Numerical calculations suggest that d n ≈ 0 . 21 / √ log n . Conjecture (B´ aez-Duarte, Balazard, Landreau, Saias (2000)) � d n ∼ C 0 / log n where � C 0 := 2 + γ − log(4 π ) = 0 . 2149 . . .

Behavior of the sequence ( d n ) Numerical calculations suggest that d n ≈ 0 . 21 / √ log n . Conjecture (B´ aez-Duarte, Balazard, Landreau, Saias (2000)) � d n ∼ C 0 / log n where � C 0 := 2 + γ − log(4 π ) = 0 . 2149 . . . Theorem (B´ aez-Duarte, Balazard, Landreau, Saias (2000)) There exists C > 0 such that d n ≥ C / √ log n for all n.

How to compute d n ? � 1 Notation: � g , h � := 0 g ( x ) h ( x ) dx .

How to compute d n ? � 1 Notation: � g , h � := 0 g ( x ) h ( x ) dx . Gram–Schmidt : There is a unique sequence of functions e 2 , e 3 , . . . on (0 , 1) such that: � e j , e k � = δ jk ∀ j , k , span { e 2 , . . . , e n } = span { f 2 , . . . , f n } ∀ n , � e k , f k � > 0 ∀ k .

How to compute d n ? � 1 Notation: � g , h � := 0 g ( x ) h ( x ) dx . Gram–Schmidt : There is a unique sequence of functions e 2 , e 3 , . . . on (0 , 1) such that: � e j , e k � = δ jk ∀ j , k , span { e 2 , . . . , e n } = span { f 2 , . . . , f n } ∀ n , � e k , f k � > 0 ∀ k . A calculation shows that, for all choices of scalars λ 2 , . . . , λ n , n n n 2 � � |� 1 , e k �| 2 + � � � | λ k − � 1 , e k �| 2 . � 1 − λ k e k = 1 − � � � k =2 k =2 k =2

How to compute d n ? � 1 Notation: � g , h � := 0 g ( x ) h ( x ) dx . Gram–Schmidt : There is a unique sequence of functions e 2 , e 3 , . . . on (0 , 1) such that: � e j , e k � = δ jk ∀ j , k , span { e 2 , . . . , e n } = span { f 2 , . . . , f n } ∀ n , � e k , f k � > 0 ∀ k . A calculation shows that, for all choices of scalars λ 2 , . . . , λ n , n n n 2 � � |� 1 , e k �| 2 + � � � | λ k − � 1 , e k �| 2 . � 1 − λ k e k = 1 − � � � k =2 k =2 k =2 Distance formula n � d 2 |� 1 , e k �| 2 . n = 1 − k =2

How to compute � 1 , e k � in practice? Four steps: Compute � 1 , f k � . This is easy since � 1 ζ ( s ) � 1 k − 1 = log k f k ( x ) x s − 1 dx = lim � � 1 , f k � = lim . s k s k s → 1 s → 1 0

How to compute � 1 , e k � in practice? Four steps: Compute � 1 , f k � . This is easy since � 1 ζ ( s ) � 1 k − 1 = log k f k ( x ) x s − 1 dx = lim � � 1 , f k � = lim . s k s k s → 1 s → 1 0 Compute � f j , f k � (see next slide).

How to compute � 1 , e k � in practice? Four steps: Compute � 1 , f k � . This is easy since � 1 ζ ( s ) � 1 k − 1 = log k f k ( x ) x s − 1 dx = lim � � 1 , f k � = lim . s k s k s → 1 s → 1 0 Compute � f j , f k � (see next slide). Deduce � e j , f k � (see next slide).

How to compute � 1 , e k � in practice? Four steps: Compute � 1 , f k � . This is easy since � 1 ζ ( s ) � 1 k − 1 = log k f k ( x ) x s − 1 dx = lim � � 1 , f k � = lim . s k s k s → 1 s → 1 0 Compute � f j , f k � (see next slide). Deduce � e j , f k � (see next slide). Compute � 1 , e j � by solving the triangular linear system n � � 1 , e j �� e j , f k � = � 1 , f k � ( k = 2 , . . . , n ) . j =2

The scalar products � f j , f k � and � e j , f k � To compute � f j , f k � , we can use Theorem (Vasyunin, 1996) Let j , k ≥ 2 , and set ω := exp(2 π i / jk ) . Then jk − 1 jk − 1 � f j , f k � = 1 � q �� q � ω − qr ( ω − r − 1) log(1 − ω r ) , � � jk j k q =1 r =1

The scalar products � f j , f k � and � e j , f k � To compute � f j , f k � , we can use Theorem (Vasyunin, 1996) Let j , k ≥ 2 , and set ω := exp(2 π i / jk ) . Then jk − 1 jk − 1 � f j , f k � = 1 � q �� q � ω − qr ( ω − r − 1) log(1 − ω r ) , � � jk j k q =1 r =1 And to obtain � e j , f k � we use Proposition Let P jk := � f j , f k � and L kj := � f k , e j � . Then L is a lower-triangular matrix and P = LL t (Cholesky decomposition).

The matrix P jk := � f j , f k � for n = 9 2 3 4 5 6 7 8 9 2 0.1733 0.1063 0.1184 0.0918 0.0931 0.0784 0.0778 0.0683 3 0.1063 0.1770 0.1220 0.1118 0.1178 0.0976 0.0908 0.0914 4 0.1184 0.1220 0.1618 0.1194 0.1103 0.1023 0.1060 0.0912 5 0.0918 0.1118 0.1194 0.1456 0.1125 0.1019 0.0956 0.0918 6 0.0931 0.1178 0.1103 0.1125 0.1313 0.1049 0.0957 0.0909 7 0.0784 0.0976 0.1023 0.1019 0.1049 0.1192 0.0976 0.0889 8 0.0778 0.0908 0.1060 0.0956 0.0957 0.0976 0.1089 0.0910 9 0.0683 0.0914 0.0912 0.0918 0.0909 0.0889 0.0910 0.1002

The matrix L kj := � f k , e j � for n = 9 2 3 4 5 6 7 8 9 2 0.4163 0 0 0 0 0 0 0 3 0.2554 0.3343 0 0 0 0 0 0 4 0.2845 0.1475 0.2430 0 0 0 0 0 5 0.2205 0.1659 0.1325 0.2277 0 0 0 0 6 0.2237 0.1814 0.0819 0.0976 0.1792 0 0 0 7 0.1883 0.1480 0.1107 0.0929 0.0991 0.1764 0 0 8 0.1868 0.1288 0.1395 0.0638 0.0721 0.0841 0.1471 0 9 0.1641 0.1479 0.0934 0.0822 0.0651 0.0664 0.0863 0.1409

Is � e j , f k � always positive? Here is what we have been able to establish.