A positivity conjecture related to the Riemann zeta function Thomas Ransford Universit´ e Laval, Quebec City, Canada New Developments in Complex Analysis and Function Theory University of Crete, July 2018
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 2002) 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.
Behavior of ( d n ) (B´ aez-Duarte et al, 2000)
Behavior of ( d n ) (Ba´ ez-Duarte et al, 2000)
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.
Is � e j , f k � always positive ? Here is what we have been able to establish. Numerical computation � e j , f k � > 0 for 2 ≤ j ≤ k ≤ 50000 .
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