Zeta Function

7/30/2019 Zeta Function

1/85

Slide 1:

Some non-standard methods

to perform calculations

with Riemanns zeta function

Yuri Matiyasevich

Steklov Institute of Mathematics

at St.Petersbourg

http://logic.pdmi.ras.ru/~yumat/personaljournal/artlessmethod

These are annotated slides of the talk given by the author on November 22, 2012 atjoin session of international meeting Zeta functions (http://www.mccme.ru/poncelet/2012zeta/) and seminar Globus (http://www.mccme.ru/ium/globus.html ). The lat-ter seminar is devoted to mathematics in general, and by this reason the talk included ashort introduction to the history of the study of Riemanns zeta function giving necessary(and, hopefully, sufficient) background for understanding the rest of the talk.

Experts may skip the first 21 slides and jump to Slide 22.Other materials related to this talk, including video, can be found at http://logic.

pdmi.ras.ru/~yumat/personaljournal/artlessmethod/talks/poncelet_2012 .

ATTENTION

The talk was based on intensive computations. After the talk was presented, GlebBeliakov from Deakin University, Australia, independently performed calculation ofthe same determinants. The values obtained by him are in good agreement with thevalues obtained by the author as long as the sizes of matrices are below 38003800.However, for larger sizes there is great discrepancy. By this reasons some originalslides were modified or removed.
http://logic.pdmi.ras.ru/~yumat/personaljournal/artlessmethod/talks/poncelet_2012http://www.mccme.ru/poncelet/2012zeta/http://www.mccme.ru/poncelet/2012zeta/http://www.mccme.ru/ium/globus.htmlhttp://logic.pdmi.ras.ru/~yumat/personaljournal/artlessmethod/talks/poncelet_2012http://logic.pdmi.ras.ru/~yumat/personaljournal/artlessmethod/talks/poncelet_2012http://logic.pdmi.ras.ru/~yumat/personaljournal/artlessmethod/talks/poncelet_2012http://logic.pdmi.ras.ru/~yumat/personaljournal/artlessmethod/talks/poncelet_2012http://logic.pdmi.ras.ru/~yumat/personaljournal/artlessmethod/talks/poncelet_2012http://www.mccme.ru/ium/globus.htmlhttp://www.mccme.ru/poncelet/2012zeta/http://www.mccme.ru/poncelet/2012zeta/


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Slide 2:

Blowing the Trumpet

The physicist George Darwin used to

say that every once in a while one

should do a completely crazy exper-

iment, like blowing the trumpet to

the tulips every morning for a month.

Probably nothing will happen, but if

something did happen, that would be

a stupendous discovery.

Ian HackingRepresenting and Intervening

Cambridge University Press

p.154, 1983

This research started as a kind of mathematical blowing the trumpet to the tulipsbut the outcome seems to be striking and deserving further investigation.

George Darwin was a son of famous naturalist Charles Darwin.


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Slide 3:

Riemanns zeta function

Dirichlet series:

(s) =

n=1

ns =

1

1s+

1

2s+

1

3s+ +

1

ns

+ . . . (1)

The series converges for s > 1 and diverges at s = 1:

1

1

+1

2

+1

3

+1

4

+ . . . (2)

The zeta function is named after Georg Friedrich Bernhard Riemann but it was pre-viously studied by Leonhard Euler.


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Slide 4:

Basel Problem (Pietro Mengoli, 1644)

1

12+

1

22+

1

32+ +

1

n2

+ = ?

Leonhard Euler:

1

12+

1

22+

1

32+ +

1

n2

+ = (2) = 1.64493406684822644 . . .(3)

2

6= 1.64493406684822644 . . .(4)

(2) =2

6(5)

Pietro Mengoli asked in 1644: What is the value of this sum?Euler calculated more than a dozen of decimal digits of the sum, which wasnt an easy

exercise because the series in (3) converges very slowly. Luckily, Euler invented what isnowadays called EulerMaclaurin summation.

Knowing the value (4), Euler conjectured the equality (5). His first proof given in1735 was not rigorous by todays standards but later he returned to this problem severaltimes and gave a number of quite rigorous proofs.


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Slide 5:

Other values of (s) found by Euler

(2) =1

62 (6)

(4) =1

904 (7)

(6) =1

9456 (8)

(8) =1

94508 (9)

(10) = 691638512875

10 (10)

(12) =2

1824322512 (11)

(14) =3617

32564156625014 (12)

Euler didnt stop by merely answering Mengolis question and found many other valuesof (s). But why such strange numerators of (10) and (14)?


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Slide 6:

Bernoulli numbers

x

ex 1=

k=0

Bkxk

k!(13)

B0 = 1, B2 =1

6, B4 =

1

30, B6 =

1

42, B8 =

1

30, (14)

B10 =5

66, B12 =

691

2730, B14 =

7

6, B16 =

3617

510(15)

B1 = 1

2, B3 = B5 = B7 = B9 = B11 = = 0 (16)

No doubts that Euler immediately recognized 691 and 3617 as numerators of the so-called Bernoulli numbers. Named after Jacob Bernoulli, these numbers can be defined inmany ways, in particular, from the coefficients in the Taylor expansion (13).

Notice that starting from 3, all Bernoulli numbers with odd indices are equal to zero.


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Slide 7:

Other values of (s) found by Euler

(6) = 1945

6 = 25

6!B6

6 (17)

(8) =1

94508 =

27

8!B8

8 (18)

(10) =691

63851287510 =

29

10!B10

10 (19)

(12) =2

1824322512 =

211

12!B12

12 (20)

(14) =

3617

32564156625014

=

213

14!B14

14

(21)

(2k) = (1)k+122k1

(2k)!B2k

2kk = 1, 2, . . . (22)

After substituting (14)(15) into (6)(12), it is easy to guess the form of the otherfactors, and Euler gave general formula (22).

We still know very little about the values of (s) for odd positive values of s.


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Slide 8:

One more value of (s) given by Euler

(0) = 10 + 20 + 30 + = 1 + 1 + 1 + . . . = 1

2(23)

(s) = (1 2 2s)(s) (24)

= (1 2 2s)(1s + 2s + 3s + 4s + . . . ) (25)

= 1s + 2s + 3s + 4s + . . .

2 2s 2 4s . . . (26)

= 1s

2s + 3s

4s + . . . (27)

The alternating Dirichlet series

1s 2s + 3s 4s + . . . (28)

converges for s > 0.

1 1 + 1 1 + 1 =1

2(29)

Euler proceeded by giving a striking value (23) for (0). His argumentation is veryimportant for understanding the main part of this talk.

Namely, Euler considered function (24) defined also by alternating Dirichlet series (27).For s = 0 the partial sums of this series oscillate around 12 and plugging this number asthe value of (0) in (24), one comes to (23).


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Slide 9:

Other values of (s) given by Euler

(0) = 10 + 20 + 30 + = 1 + 1 + 1 + = 1

2(30)

(1) = 11 + 21 + 31 + = 1 + 2 + 3 + . . . = 1

12(31)

(2) = 12 + 22 + 32 + = 1 + 4 + 9 + = 0 (32)

(3) = 13 + 23 + 33 + = 1 + 8 + 27 + . . . =1

120(33)

..

.

..

.

..

.(11) = 111 + 211 + = 1 + 2048 + . . . =

691

32760(34)

(m) = Bm+1

m + 1m = 0, 1, . . . (35)

(2) = (4) = (6) = = 0 (36)

In similar style Euler obtain other values for negative integer argument. The ap-pearance of 691 in the numerator of (11) suggests that Bernoulli numbers are againinvolved, and Euler gave the general formula (35).

In particular, recalling (16) he got (36).


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Slide 10:

The functional equation

(2k) = (1)k+122k1

(2k)!B2k

2kk = 1, 2, . . . (37)

(m) = Bm+1

m + 1m = 0, 1, . . . (38)

m = 2k

1 (39)

(1 2k) = (1)k212k2k(2k 1)!(2k) k = 1, 2, . . . (40)

Selecting m according to (39), one can eliminate Bernoulli numbers from (37) and (38)and get (40).


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Slide 11:

Euler Identity Fundamental Theorem of Arithmetic

Theorem [Euler].

(s) =1

1s+

1

2s+

1

3s+ +

1

ns+ . . . (41)

=

p is a prime

1

1 ps(42)

Proof.

p is a prime

1

1 ps=

p is a prime

(1 +1

ps+

1

p2s+

1

p3s+ . . . ) (43)

=1

1s+

1

2s+

1

3s+ +

1

ns+ . . . (44)

= (s) (45)

In order to obtain (43) from the right-hand side, one needs just to sum up the geo-metrical progression. Obtaining (43) from (44) is based on the Fundamental Theorem ofArithmetic saying that every natural numbers can be represented as product of powers ofdifferent primes, and this can be done in a unique way (up to the order of factors).


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Slide 12:


Dirichlet series:

(s) =

n=1ns

s = + it

5 10 15 20

6

4

2

2

4

6

The series converges in the semiplane Re(s) > 1 and defines a functionthat can be analytically extended to the entire complex plane except forthe point s = 1, its only (and simple) pole.

Riemann began to study zeta-function for complex argument and established the tra-dition to denote it as s = + it.


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Slide 13:


Eulers values of (s) for negative integer s were correct:

(m) = Bm+1

m + 1m = 0, 1, . . . (46)

(2) = (4) = (6) = = 0 (47)

Points s1 = 2, s2 = 4, . . . , sk = 2k, . . . are called the trivial zeroesof the zeta function, and it has no other real zeroes.

Great Euler was right!


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Slide 14:

The functional equation

Euler:

(1 2k) = (1)k212k2k(2k 1)!(2k) k = 1, 2, . . . (48)

Riemann:

(1 s) = coss

2

21ss(s)(s) s = + it (49)

To pass from (48) to (49) one had to guess correct counterpart of (1)k.


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Slide 15:

Function (s)

(1 s) = coss2

21ss(s)(s) (50)

coss2

=

( s2

+ 12

)( s2

+ 12

)(s) =

2s1

s

2

s

2+ 1

2

(51)

1s

2 (s) 1s2

+ 1 (1 s) (1s)

= s

2 (s 1) s2

+ 1 (s) (s)

(52)

(1 s) = (s) (53)

Function (s) is entire, its zeroes are exactly non-trivial (i.e., non-real)zeroes of(s).

In the right-hand side of (52), the pole of zeta-function is cancelled by the factor s1,and the poles of the -factor are cancelled by the trivial zeroes of the zeta-function.


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Slide 16:

Function (t)

(s) = s

2 (s 1)

s

2+ 1

(s)

g(s)

(54)

s =1

2+ it (55)

(t) = 1

2+ it

=

1

1

2+ it

= (t) (56)

(t) is an even entire function with real value for real values of itsargument t.

The Riemann Hypothesis. All zeroes of (t) are real numbers.

Making the change of the variable (55) is again a tradition due to Riemann. However,for functions we use modern notation introduced by E. Landau, Riemann himself calledour (t) as (t).

There is no standard notation for the important factor in (54) that well be denot-ing g(s).

The Riemann Hypothesis was originally stated in terms of function (t), not in termsof function (s).


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Slide 17:

Chebyshev function (x)

(x) = ln(LCM(1, 2, ..., x)) (57)

=

qxq is a power

of prime p

ln(p) (58)

=

qx

q=1(q) (59)

Von Mangoldt function

(q) =

ln(p), q = pk for some prime p

0, otherwise(60)

Riemann established a relationship between prime numbers and zeroes of the zetafunction by giving a (somewhat complicated) expression for (x) the number of primesbelow x via certain sums over these zeroes. Chebyshev function (x) turned out to bemore convenient for studding the distribution of prime numbers.


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Slide 18:

Logarithmic derivative of (s)

(s) =

p is a prime

1

1 ps(61)

(s)

(s)=

p is a prime

ln

1

1 ps

(62)

=

p is a prime

ln(p)

1 ps(63)

=

p is a prime

ln(p)k=1

pks (64)

=

n=1

(n)ns (65)

Von Mangoldt function plays important role in the theory of the zeta-functions becausethe values of the former functions are nothing else but (negations of) the coefficients inthe Dirichlet series of the logarithmic derivative of the latter function.


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Slide 19:

Chebyshev function (x)

0 5 10 15 20

5

10

15

20

The jumps occur at primes and at their powers.


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Slide 20:

Von Mangoldt Theorem

(x) = ln(LCM(1, 2, ...,x)) =x

q=1

(q) (66)

Theorem (Hans Carl Fridrich von Mangoldt [1895]). For anynon-integer x > 1

(x) = x

()=0

x

n

x2n

2n ln(2) (67)

Theorem (Jacques Salomon Hadamard [1893]).

(s) = (0)

()=0

1

s

(68)

(s) = g(s)(s) = g(s)

p is a prime

1

1 ps(69)

The proof of von Mangoldt theorem is based on the appearance of the two productsin (68) and (69).


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Slide 21:

Von Mangoldt Theorem

(x) x

||


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Slide 22:

Blowing the Trumpet

Assuming that all zeroes of(t) are real and simple, let them be denoted1,2, . . . with 0 < 1 < 2 < . . . , thus the non-trivial zeroes of (s)are 1

2 i1,

1

2 i2, . . .

1 2 3 5 6712356 7

1 = 14.1347 . . . 2 = 21.0220 . . . (74)

3 = 25.0109 . . . 4 = 30.4249 . . . (75)

Suppose that we have found 1, 2, . . . , N1; how could these numbersbe used for calculating an (approximate) value of the next zero N?

It was a rather strange idea to seek an answer to this question because known initialzeroes are distributed rather irregularly.


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Slide 23:

Interpolating determinant

Let us try to approximate (t) by some simpler function also having zeroesat the points 1, . . . ,N1.

Consider an interpolating determinant with even functions f1, f2, . . .

f1(1) . . . f1(N1) f1(t)...

. . ....

...fN(1) . . . fN(N1) fN(t)

(76)

Selecting fn(t) = t2(n1)

we would obtain just an interpolating polynomial

C

N1

n=1

(t2 n2) (77)

having no other zeros.

Clearly, the determinant (76) vanishes for t = 1, . . . ,N1 because for such valuesof t the matrix contains two equal columns.


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Slide 24:

Modelling situation

If 1 , 2 , . . . are zeros of the function

(t) =N

k=1

fk(t) (78)

then the determinant

f1(

1) . . . f1(

N1) f1(t)

.... . .

......

fN(

1) . . . fN(

N1) fN(t)

(79)

vanishes at every zero of (t).

The determinant (79) vanishes as soon as t is equal to any zero of (t) because forsuch a t the rows of the matrix sum up to the zero row.


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Slide 25:

Selecting f1, f2, . . .

(t) =

n=1

n(t) n(t) = g1

2+ it

n( 1

2+it) (80)

(t) =

n=1

n(t) for Im(t) < 1

2(81)

(t) = (

t) =

n=1

n(

t) for Im(t) >1

2 (82)

(t) =

n=1

n(t) + n(t)

2=

n=1

n(t) (83)

Our case differs from the modelling one in two respects. First, the sum in (80) isinfinite. Second, its summands arent even functions.

To remove the first difference we introduce even functions n(t). However, the equal-ity (83) is purely symbolic because the half planes of convergence of the two series (81)and (82) dont intersect.


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Slide 26:

Our interpolating determinant

N(t) =

1(1) . . . 1(N1) 1(t)...

. . ....

...N(1) . . . N(N1) N(t)

(84)

n(t) =g1

2 it

n( 1

2it) + g

1

2+ it

n( 1

2+it)

2(85)

g(s) = s

2 (s 1)s

2+ 1

(86)

(t) =

n=1

n(t) (87)

For determinant we need only finitely many functions, so for the moment we can ignorethe divergence of (87).


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Slide 27:

Blowing the trumpet: the outcome

47(138.11604)=2.18497 . . . 101216 < 0

47 = 138.1160420545334. . .47(138.11605)=+4.68242 . . . 10

1216 > 0

321(572.419984132452764045276927)=1.75211 . . . 1034969 < 0

321 = 572.41998413245276404527692710734686. . .321(572.419984132452764045276928)=+5.72682 . . . 10

34968 > 0

600(939.02430089921838218421841870917956106917\

858909398785769363406563371 = +3.90403 . . . 10108715 > 0

600 = 939.02430089921838218421841870917956106917\

85890939878576936340656337113493189 . . .

600(939.02430089921838218421841870917956106917\

858909398785769363406563372) = 9.67471 . . . 10108716 < 0

Here are just a few numerical examples.A zero of 47(t) has 8 decimal digits coinciding with digits of 47.For N = 321 there are already 15 coinciding decimal digits.For N = 600 the number of common digits increases to 38.


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Slide 28:


47(138.11604)=2.18497 . . . 101216 < 0

47 = 138.1160420545334. . .47(138.11605)=+4.68242 . . . 10

1216 > 0

47(139.7362)=+1.27744 . . . 101216 > 0

48 = 139.736208952121. . .47(139.7363)=9.88309 . . . 10

1216 < 0

47(141.12370)=1.85988 . . . 101217 < 0

49 = 141.1237074040211. . .47(141.12371)=+2.40777 . . . 10

1217 > 0

47(143.11184)=+8.00594 . . . 101218 > 0

50 = 143.1118458076206. . .47(143.11185)=8.98353 . . . 10

1218 < 0

It turned out that N(t) allows us to calculate good approximations not only to thenext not yet used zero N but to N+k as well for values of k that are not too large.


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Slide 29:


321(572.419984132452764045276927)=1.75211 . . . 1034969 < 0

321 = 572.41998413245276404527692710734686. . .321(572.419984132452764045276928)=+5.72682 . . . 10

34968 > 0

321(573.61461052675812998724603)=+2.27756 . . . 1034967 > 0

322 = 573.6146105267581299872460321096433. . .321(573.61461052675812998724604)=9.27805 . . . 10

34967 < 0

321(575.09388601449488560745717)=5.83758 . . .

1034968

< 0323 = 575.0938860144948856074571722821982. . .321(575.09388601449488560745718)=+2.78750 . . . 10

34967 > 0...............................................................................................................

321(585.98456320498830057418)=5.42229 . . . 1034968 < 0

331 = 585.9845632049883005741872589855. . .321(585.98456320498830057419)=+8.57284 . . . 10

34969 > 0

These are just a few examples, tables showing the number of digits common to(N + k)th zero of (t) and a zero of N(t) for diverse N and k can be found at

http://logic.pdmi.ras.ru/~yumat/personaljournal/artlessmethod .
http://logic.pdmi.ras.ru/~yumat/personaljournal/artlessmethodhttp://logic.pdmi.ras.ru/~yumat/personaljournal/artlessmethod


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Slide 30:

Slide removed

I dont have a full explanation of the high accuracy of zeroes of determinants N(t)as approximations to zeroes of (t). Two heuristic hints will be presented now.


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Slide 31:

Partial explanation. I

(t) =

n=1

n(t) (88)

N(t) =

1(1) . . . 1(N1) 1(t)...

. . ....

...

N(1) . . . N(N1) N(t)

=

N

n=1

N,nn(t) (89)

N,n = (1)N+n

1(1) . . . 1(N1)

... . . . ...n1(1) . . . n1(N1)

n+1(1) . . . n+1(N1)...

. . ....

N(1) . . . N(N1)

(90)

The s are particular numbers defined via the zeroes of the zeta-function.


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Slide 32:

Normalization

(t) =

n=1

n(t) (91)

N(t) =

1(1) . . . 1(N1) 1(t)...

. . ....

...N(1) . . . N(N1) N(t)

=

N

n=1

N,nn(t) (92)

N,n =N,n

N,1N,1 = 1 (93)

N(t) =N

n=1

N,1N,nn(t) = N,1

N

n=1

N,nn(t) (94)

The normalization doesnt influence zeroes, that is, N(t) has the same zeroes asNn=1 N,nn(t).


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Slide 33:

Normalized coefficients 47,n

47

n=1

47,nn(t) (95)

We the sum (95) is not a sharp but a smooth truncation of the divergent series (87).It is known that smooth truncation can accelerate convergence of a series and it can eventransform a divergent series into a convergent one.

The smoothness of the truncation in (95) might be the first reason why the summandsof the divergence series (87) are useful for calculation of the zeroes.


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Slide 34:

Normalized coefficients 47,n with logarithmic scale

47,n 1+ 47 log(n) (96)

The curve from Slide 33 on which the coefficients 47,n lie looks like a logarithmiccurve.

This is better seen on the plot of the same coefficients but with logarithmic scale.Why do the initial coefficients lie approximately on a straight line?


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Slide 35:


48,n 1+ 48 log(n) (97)

The graphs for N = 48, 49, 50, 51 look very similar to the graph for N = 47 fromSlide 34.


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Slide 36:


52,n 1+ 52 log(n) (101)

Normalized coefficients 52,n show slightly different behaviour of trailing coefficients.


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Slide 37:


54,n 1+ 54 log(n) (103)

What a difference with previous graphs! Now most of the coefficients are negative andgreater than 1 in absolute value. Nevertheless, the initial coefficients do lie on a straightline and 54(t) gives good predictions for a few of the next zeroes 54, 55, . . .


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Slide 38:


55,n 1+ 55 log(n) (104)

PictuGrapfs for N = 55, 56, 57 look similar to the graph for N = 47. Case N = 54demonstrates sporadic behavour. Why number 54 is special?


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Slide 39:


130,n 1 + 130,2dom2(n) + 130 log(n) (107)

domm(k) =

1, ifm | k0, otherwise

(108)

We skip many values ofN (the skipped and many other graphs can be found at http://logic.pdmi.ras.ru/~yumat/personaljournal/artlessmethod ) and examine a newphenomenon in the case N = 130.

Now the initial coefficients lie on two parallel lines rather than on a single line, soinstead of an analog of (96), (97), (101), (103) and (104), in the case N = 130 we shoulduse approximation (107).
http://logic.pdmi.ras.ru/~yumat/personaljournal/artlessmethodhttp://logic.pdmi.ras.ru/~yumat/personaljournal/artlessmethodhttp://logic.pdmi.ras.ru/~yumat/personaljournal/artlessmethodhttp://logic.pdmi.ras.ru/~yumat/personaljournal/artlessmethod


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Slide 40:


131,n 1 + 131,2dom2(n) + 131 log(n) (109)

Graph for N = 131 shows that the case N = 130 is sporadic.


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Slide 41:


169,n 1 + 169,2dom2(n) + 169 log(n) (111)

But with increasing N splitting into two lines becomes typical.


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Slide 42:


321,n (1)n1 (113)

1 + 321,2dom2(n) with 321,2 2 (114)

Skipping again many pictugraphs, we consider new pattern exhibited for N = 321 now initial coefficients alternate between approximately +1 and 1.


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Slide 44:

Functions N(s)

(s) =n=1

(1)n1ns (119)

N(s) =N

n=1

N,nns (120)

N(t) =1

2N,1g12 it N 12 it+ g12 + it N 12 + it (121)

N(s)?

(s) = (1 2 2s)(s) (122)

(s)?

N(s)

1 2 2s(123)

Based on (113), we introduce functions N(s) mimicking Eulers function (s).Shouldnt (122) be true? If yes, than well be able to calculate (s) via N(s).


45/85

Slide 45:

Calculation of (s)

For s = 12

+ 100i

(s)321(s)122s

1

= +1.34... 1012 (124)

(s)600(s)122s

1

= 2.95... 1024 (125)

For s = 14

+ 100i

(s)600(s)

122s

1

= 2.64... 1024 (126)

(s)3000(s)122s

1

= +1.19... 10125 (127)

For s = 14

+ 1000i

(s)3000(s)122s

1

= +4.07... 10126 (128)

Indeed, such calculations turned out to be possible. For N = 321 we get 12 correctdecimal digits of(s), and 24 correct decimal digits in the case N = 600. However, earlier(Slides 27 and Slide 29) we got for the same values of N much more correct decimal digitsof the s. How is it possible?


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Slide 46:


Let us have a closer look at the areas marked in yellow.


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Slide 48:

Normalized coefficients: general case

N,n

M

m=1N,mdomm(n) + N log(n) (132)

N,n =

n

m=1N,mdomm(n) =

m|n N,m n = 1, 2, . . . (133)

N,n =m|n

n

m

N,n (134)

Mobius function

(k) =

(1)m, ifk is the product ofm different primes

0, otherwise(135)

1

(s)=

1n=1

ns=

n=1

(n)nsn=1

ns

n=1

(n)ns = 1 (136)

In fact, there are similar even smaller splitting depending on the values of n modulo4, 5, . . . , so we need to use (132).

To give formal definition of N,n we ignore the logarithmic term in (132) and demandthat it to be exact equality when M = n. Resulting triangular linear system (133) hassolution (134).

The Mobius function also play important role in the theory of the zeta-functions dueto (136).


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Slide 49:

Function N,M(s)

M 3000,M1 12 2 4.9... 10126

3 7.40565... 10126

4 2.85890... 10284

5 1.85782... 10411

6 2.56167... 10509

7 5.02202... 10585

8 4.39701... 10641

9 1.08444...

10681

10 1.90599... 10716

N(s)

(s)=

N

n=1 N,nns

n=1 ns

(137)

=n=1

N,nns (138)

N(s) =

n=1 N,nns

(s) (139)

N(s) (1 2 2s)(s) N(s)

Mn=1

N,nns

(s)

N,M(s)

(s) (140)

Equivalently, numbers N,n can be defined as the coefficients in the ratio (137).The small numerical values of 3000,n imply that between the initial 3000,n there is a

number of almost linear relations of the form (134).To what extent can function N,M(s) play the role of Eulers factor 1 2 2

s?


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Slide 50:

Improved calculation of(s) for s = 14+ 1000i

N(s) N,M(s)(s) N,M(s) =M

n=1N,nn

s (141)

M 3000,M

N(s)3000,M(s)(s) 1

2 2 4.9... 10126 2.46329... 10126

3 7.40565... 10126 8.84947... 10285

4 2.85890... 10284 5.43870... 104125 1.85782... 10411 7.16503... 10

510

6 2.56167... 10509 1.35156... 10585

7 5.02202... 10585 1.14450... 10641

8 4.39701... 10641 2.13729... 10681

9 1.08444... 10681 2.18430... 10681

10 1.90599... 10716 2.18430... 10681

11 2.37291... 10681 2.56047... 10681

For M = 2 we have almost Euler-like approximation (123).Increasing M up to 8 greatly improves the accuracy.But further increasing the value of M doesnt lead to better approximations. The

reason is as follows: decreasing of (absolute values of) 3000,M stops at M = 10.


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Slide 51:

log10(|3000,n|), magenta, if3000,n > 0, green otherwise

For n = 2..10 the absolute values of 3000,n decrease, and their logarithms lie (why?)on a smooth curve in spite of the fact that the values themselves have different signs.

Starting from n = 11 the absolute values of 3000,n oscillate between approximately10681 and 10752. The values of n corresponding to the first of these two cases are allprimes or powers of primes, and corresponding values of 3000,n are all positive.


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Slide 52:

3000,n, magenta, if3000,n > 0, green otherwise

Closer look at values of 3000,n for prime and prime power values of n reveals finerstructure.


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Slide 53:

3000,n

ln(n), magenta, if3000,n > 0, green otherwise

This structure becomes more transparent when we normalize the s by ln(n).


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Slide 54:

3000,n(n)

3000 =3000,13ln(13)

= 9.895811... 10682 (142)

3000,pk/ ln(p)

3000 1

< 3.85... 1073 for 13 pk 419, p is a prime

3000,pk

ln(p) 3000

< 3.81... 10754 for 13 pk 419, p is a prime

It is even better to normalize by von Mangoldts (n) (defined on Slide 17).For a long range of prime and prime power values of n the value of the ratio

3000,n(n)

remains almost constant. For non-prime-power values of n in the same range the val-ues of |3000,n| are by more than 70 decimal orders smaller than 3000, that is, 3000,napproximates very well (a multiple of) von Mangoldt function (n).

The connections between zeta-function and prime numbers is usually attributed toEuler product (42), however, in this case it isnt clear how to establish connection betweenEuler product and the values of 3000,n in this range.

Another intriguing question is: why the values of 3000,n for a few initial values of nare so different? Their structure was analyzed to some extent in

http://logic.pdmi.ras.ru/~yumat/personaljournal/artlessmethod/

talks/leicester2012/leicester_2012_full.pdf .
http://logic.pdmi.ras.ru/~yumat/personaljournal/artlessmethod/talks/leicester2012/leicester_2012_full.pdfhttp://logic.pdmi.ras.ru/~yumat/personaljournal/artlessmethod/talks/leicester2012/leicester_2012_full.pdfhttp://logic.pdmi.ras.ru/~yumat/personaljournal/artlessmethod/talks/leicester2012/leicester_2012_full.pdfhttp://logic.pdmi.ras.ru/~yumat/personaljournal/artlessmethod/talks/leicester2012/leicester_2012_full.pdf


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Slide 55:

3000(s)

(s)=

n=1

3000,nns (143)

This is just a definition of numbers 3000,n (case N = 3000 of (138)).


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Slide 56:

3000(s)

(s)

8

n=1

3000,nns

Filling in the table on Slide 50, we dropped all summands in (143) but a few initials,say, the first eight.


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Slide 57:

3000(s)

(s)

11

n=1

3000,nns

Let us now do the opposite drop initial 11 summands from (143)


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Slide 59:

Calculating zeta derivative at zeros

3000(s)

(s)

11n=1

3000,nns

?

3000(s)

(s)(147)

3000(s)

11

n=1

3000,nns

(s)

?

3000(s) (148)

3000

1

2+ ik

?

3000

1

2+ ik

(149)

30001

2+ i100

3000

1

2+ i100

1 = 1.024... 1036 (150) 3000

1

2+ i500

3000

1

2+ i500

1 = 2.786... 1074 (151)

Let us check our guess at first at a pole of (s) when (148) simplifies to (149). Indeed,we get a lot of correct digits. This is rather surprising by two reasons. First, we calculateoutside the semiplane Re(s) > 1 of the convergence of (65). Second, passing from (147)to (148) we multiplied both sides by (s) but in our case its value is zero.


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Slide 60:

Calculating zeta derivative

3000(s)

11

n=1

3000,nns

(s)

?

3000(s) (152)

s =1

4+ 1000i (153)

3000 (s)

11

n=13000,nn

s

(s)

3000 (s) 1

= 6.44... 1073 (154)

We can get many digits of (s) for s different from a pole as well. However, for thiswe need to know the value of (s) with much higher accuracy than that from table onSlide 50.


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Slide 62:

Calculating both zeta and its derivative. II

3000(s)

11

n=1

3000,nns

(s) 3000

(s) (157)

3000(1 s)

11

n=1

3000,nns1

(1 s) 3000

(1 s) (158)

g(s)(s) = g(1 s)(1 s) (159)

g(s)(s) + g(s)(s) = g(1 s)(1 s) g(1 s)(1 s) (160)

Solving this system for s= 14+ 1000i produces 752 correct decimal digits

for (s) and 72 correct decimal digits for (s).

Another way is to use the functional equation relating the values of (s) and (s)with values of (1 s) and (1 s).


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Slide 63:

Near the pole

(s) = (1 2 2s)(s) (161)

N(s) N,M(s)(s) N,M(s) =M

n=1

N,nns (162)

M 3000,M 3000,M(1)

2 2 4.9... 10126 2.46855... 10126

3 7.40566... 10126 7.14726... 10285

4 2.85891... 10284 3.71565... 10412

5 1.85782... 10411 4.26945... 10510

6 2.56167... 10509 7.17431... 10586

7 5.02202... 10585 5.49626... 10642

8 4.39701... 10641 5.73467... 10681

9 1.08444... 10681 5.61417... 10681

10 1.90599... 10716 5.61417... 10681

11 2.37291... 10681 5.39845... 10681

The role played by Eulers factor 1 2 2s in (161) was to cancel the pole of the zetafunction. To what extent our factor N,M(s) plays this role?

The small values of the s imply almost linear relation between the s, the coefficientsbeing rational numbers with small denominators.


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Slide 64:

Near the pole

(s) = (1 2 2s)(s) (161)

N(s) N,M(s)(s) N,M(s) =Mn=1

N,nns (162)

(1) = lims1

(1 2 2s)(s) (163)

= lims1

(1 2 2s) 1s 1

+ . . . (164)= (1 2 2s)

s=1

(165)

N(1)

N,M(1) (166)

3000(1)

3000,9

(1)= 1 4.1515... 10680 (167)

In spite of the fact that function 3000,M(s) doesnt vanish as s = 1, its derivativeapproximates 3000(1) very well.


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Slide 65:

log10(|3000,n|) (repeated)

We have analyzed the fine structure of numbers N,n for prime and prime power valuesof n. Is there any interesting pattern for other values of n? For example, for such n thevalue of N,n can be both positive and negative, and the sign depends on both n and N.Is there any simple rule for determining the signs?

Clearly, 3000,n cannot distinguish prime powers from the rest of integers for all n,


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Slide 66:


so let us look at larger values of n.


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Slide 67:


While after n = 420 the absolute values of 3000,n for non-prime-power values of nbegin to grow up, for prime power values of n the values of 3000,n continue to stay closeto 10681. We can clearly see points for n = 512, 625, 729.

Why the relatively slow growth of |3000,n| suddenly changes to much faster one aftern = 791?

Why after n = 797 there is no distinction between prime power and non-powers?


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Slide 68:



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Slide 69:


3000,n (3000)(n) (x) =

x

n=1

(n) (168)

Could we use the s for studding the growth of (x)? To this end we would need toestimate certain sums with N,n.


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Slide 70:

N,M(s)

N,M(s) =M

n=1

N,nns N,n =

m|n

n

m

N,n (169)

N,n =N,n

N,1N,n = (1)

N+n

1(1) . . . 1(N1)...

. . ....

n1(1) . . . n1(N1)n+1(1) . . . n+1(N1)

..

.

. ..

..

.N(1) . . . N(N1)

(170)

n(t) = g1

2 it

n( 1

2it) + g

1

2+ it

n( 1

2+it) (171)

g(s) = s

2 (s 1)s

2+ 1

0 < 1 < 2 < . . . (172)

Can we expect interesting properties of individual values of numbers N,M(s) definedin this involved way?


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Slide 71:

N,N(0)

N,M(s) =

M

n=1

N,nns (173)

N,M(0) =M

n=1

N,n (174)

3000,3000(0) =

1

2 + 1.23...10126

3001,3001(0) = 3

2+ 6.67... 10126

(175)

6000,6000(0) = 1

2 5.48... 10166

6001,6001(0) = 3

2 3.43... 10166

(176)

Numbers N,N(0) seem to be close to 1 +(1)N

2.


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Slide 72:

N,N(1)

N,M(s) =

M

n=1

N,nns

N,M(1) =

M

n=1

N,n

n(177)

N = 2520 = LCM(1, 2, 3, 4, 5, 6, 7, 8, 9, 10) (178)

2N N,N(1) = 0.9998015873172093... (179)

=

1

1 + 15039+

1

2520+1

1680+1

1260+1

1008+1

840+1

720+1

630+1

560+1

504+1

...

(180)

5039 = 2N 1, 2520 =2N

2, 1680 =

2N

3, 1260 =

2N

4, 1008 =

2N

5,

840 =2N

6, 720 =

2N

7, 630 =

2N

8, 560 =

2N

9, 504 =

2N

10

In all other examples the choice of N wasnt crucial, neighbouring values would suitequally well. Here neither N = 2519 nor N = 2521 would do, it is important that 2 2520is divisible by 2,3,. . . ,10. Similar, the pattern doesnt continue because 2 2520 isntdivisible by 11.

Also, this is the only place where the normalization (93) is crucial.


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Slide 73:

N,N(1)

(N) =1

2N

1

1 +1

2N 1 +1

2N

2+

1

2N

3+

1

2N

4+

1

2N5 + 1

2N

6+

. . .

(181)

=1

2

N

2+ 1

1

2

N + 1

2

(182)

(z) =(z)

(z)(183)

The equivalence of the two definitions of (N) was established by comparing a fewinitial terms in the asymptotic expansions, I dont have a rigorous proof.


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Slide 74:

N,N(1)

N,M(s) =

Mn=1

N,nns N,M(1) =

Mn=1

N,n

n(184)

(N) =1

2

N

2+ 1

1

2

N + 1

2

(z) =

(z)

(z)(185)

3000,3000(1)

(3000) = 1

2.46827839...

10126

(186)

6000,6000(1)

(6000)= 1 + 1.09736771... 10165 (187)


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Slide 75:


This graph looks similar to the case N = 321 on Slide 42.


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Slide 77:


Again similar graph.


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Slide 78:


This is a sporadic graph.


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Slide 79:


And this graph is sporadic.


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Slide 80:


Notice multiples change of sign.


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Slides 81-87 were removed.

Slide 88:


The s show two typical behaviour:

logarithmic for small" N like on Slide 33,

parallel for medium N like on Slide 42,

The s behave more uniformly. Already for N as small as 47 we can distinguish primesand their powers.


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Slide 89:


Here we can see n = 25, 27, 32, 49, 64, 81, 121, 125, 128.


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Slide 90:



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Slides 9195 were removed.

I found very interesting continuation of calculations for N > 12000 but this exceedsthe computational resources I have at my disposal at present, international cooperationseems to be necessary.

Also it is very interesting to see if there are similar phenomena for other L-functions.We saw that for Riemann zeta function many things are connected with its pole, so,probably, we should deal not with individual Dirichlet L-functions but with the productof all of them corresponding to fixed prime p.

The determinants N(t) were defined under the assumptions that all zeroes satisfyRiemanns Hypothesis and are simple. What would happen if Riemanns Hypothesis isvalid but there are multiple zeroes? It might completely spoil considered above methodsfor calculating the values of the zeta function, its derivative and zeroes in the case oflarge N. More likely, it will only be necessary to modify the definitions (84) and (90) byreplacing columns with repeated values of zeroes by columns with derivatives of n. Butmaybe there is no need in such modification and proving it would give a proof (under

RH) of the simplicity of all zeroes.Possible non-real zero of (t) wouldnt spoil definitions (84) and (90) but would

require making a decision which of the two zeroes, or , should be used first. WhenN is such that odd number of non-real zero of (t) are used in (90), the values of N,nand N,n will become in general non-real making (n)-like behaviour of N,n doubtful.Does this observation open a way to proving RH?


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Slide 96:

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