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CONNECTION BERNOULLI NUMBERS B N AND RIEMANN
ς(s) ZETA FUNCTION WITH ITS ZEROS
Ing. Pier Franz Roggero, Dott. Michele Nardelli, P.A. Francesco Di Noto
Abstract:
In this paper we focus attention on a relationship between the denominators
of Bernoulli numbers Bn and prime numbers.
We can define the Bernoulli's function as the analytic continuation of the
Bernoulli's formula in the field of complex numbers.
So we find an interesting correlation on the Riemann ς (s) zeta function and
the Bernoulli numbers in its zeros.
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1. BERNOULLI NUMBERs ....................................................................................................... 3 1.1 TABLE OF FACTORIZATION OF DENOMINATOR BERMOULLI NUMBERS AND
ζ(1−n) IN CORRELATION WITH PRIME NUMBERS ..................................................... 7 1.1.1 CALCULATION OF den(BN) ................................................................................ 11 1.1.2 NUMERATOR OF BN num(BN) ............................................................................ 14 1.1.3 FINDING A PROBABLY PRIME NUMBER p LARGE AS DESIRED TO 60% ....... 18 1.2 CONNECTION ζ(s) ZETA RIEMANN FUNCTION WITH PRIME NUMBERS .......... 21 1.3.1 SPECIFIC VALUES ............................................................................................. 24
2. BERNOULLI'S FORMULA EXTENDED IN THE FIELD OF REAL NUMBERS xεR.............. 27 3. BERNOULLI's FORMULA EXTENDED IN THE FIELD OF COMPLEX NUMBERS zεC, z=σ ±it
.............................................................................................................................................. 34 4. ZEROS ζ(½ ± ix)=0 ON THE CRITICAL LINE ..................................................................... 37
4.1 GRAPHICS OF THE REAL AND IMAGINARY ς(p) ................................................ 47 5. A NEW REPRESENTATION OF THE BERNOULLI FUNCTION CONNECTED WITH THE
RIEMANN ZETA FUNCTION................................................................................................. 50 5.1 A REPRESENTATION OF THE ABSOLUTE VALUES OF BERNOULLI NUMBERS
CONNECTED WITH THE RIEMANN ZETA FUNCTION FOR n≥1 ............................... 53 6. REFERENCES .................................................................................................................... 55
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1. BERNOULLI NUMBERS
The Bernoulli numbers Bn are a sequence of rational numbers with deep connections to
the number theory.
The values of the first few Bernoulli numbers are
B0=1, B1=±1 ⁄ 2, B2=1 ⁄ 6, B3=0, B4=−1 ⁄ 30, B5=0, B6=
1 ⁄ 42, B7=0, B8=−1 ⁄ 30.
If the convention B1=−1 ⁄ 2 is used, this sequence is also known as the first Bernoulli
numbers, with the convention B1=+1 ⁄ 2 is known as the second Bernoulli numbers.
Except for this one difference, the first and second Bernoulli numbers agree. Since Bn=0
for all odd n>1, and many formulas only involve even-index Bernoulli numbers, some
authors write Bn instead of B2n.
For m, n ≥ 0 we define
( ) ∑=
+++==n
k
mmmm
m nk nS 1
...21 .
This expression can always be rewritten as a polynomial in n of degree m + 1. The
coefficients of these polynomials are related to the Bernoulli numbers by Bernoulli's
formula:
( ) ∑=
−+
+
+=
m
k
k m
k m n Bk
m
mnS
0
11
1
1,
where the convention B1=+1/2 is used. (
+
k
m 1denotes the binomial coefficient, m+1
choose k .)
For example, taking m to be 1 gives the triangular numbers 0, 1, 3, 6,....
( ) ( )nnn Bn Bn +=+=+++21
1
2
02
12
2
1...21 .
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Taking m to be 2 gives the square pyramidal numbers 0, 1, 5, 14,....
Some authors use the convention B1=−1/2 and state Bernoulli's formula in this way:
( ) ( )∑=
−+
+−
+=
m
k
k m
k
k
m n Bk
m
mnS
0
11
11
1.
The Riemann zeta function ζ (s) is a function of a complex variable s=σ +it .
The following infinite series converges for all complex numbers s with real part greaterthan 1, and defines ζ (s) in this case:
( ) ∑∞
=
−+++==
1
...3
1
2
1
1
1
nsss
snsζ ( ) 1>= sR σ .
There is a relationship between the denominators of Bernoulli numbers Bn and prime
numbers
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We can define:
Denominator( Bn) = den( Bn) → n+1
den( B n ) = last factor of the factorization( B n ) if the last number is n +1 we have thatit is a prime number p
Example:
n=12
den(B12)= 2730
Denominator( B12)= last factor of the factorization( B12) = 2730 = 2 * 3 * 5 * 7 * 13 → 13 prime number.
When n odd integers we have that all the values of Bn are zero and so we find all
the n+1 even integer.
When we have a den(Bn ) that repeats again we will have all the odd integers that
aren’t prime numbers.
Examples:
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den(B8)= 30 already appeared with den(B4) →9 (not a prime number)
den(B14)= 6 already appeared with den(B2) → 15 (not a prime number)
den(B24)= 2730 already appeared with den(B12) → 25 (not prime)
den(B26)= 6 already appeared with den(B2) → 27 (not prime)
den(B32)= 510 already appeared with den(B16) → 33 (not prime)
den(B34)= 6 already appeared with den(B2) → 35 (not prime)
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1.1 TABLE OF FACTORIZATION OF DENOMINATOR BERNOULLI NUMBERS
AND ζ (1−n) IN CORRELATION WITH PRIME NUMBERS
TAB. 1
den(Bn) Factorization n+1 den (ς(1-n))
1 2 2 2 2
2 6 2 * 3 3 124 30 2 * 3 * 5 5 120
6 42 2 * 3 * 7 7 252
8 30 2 * 3 * 5 9 240
10 66 2 * 3 * 11 11 660
12 2730 2 * 3 * 5 * 7 * 13 13 32760
14 6 2 * 3 15 84
16 510 2 * 3 * 5 * 17 17 8160
18 798 2 * 3 * 7 * 19 19 14364
20 330 2 * 3 * 5 * 11 21 6600
22 138 2 * 3 * 23 23 3036
24 2730 2 * 3 * 5 * 7 * 13 25 65520
26 6 2 * 3 27 156
28 870 2 * 3 * 5 * 29 29 24360
30 14322 2 * 3 * 7 * 11 * 31 31 429660
32 510 2 * 3 * 5 * 17 33 16320
34 6 2 * 3 35 204
36 1919190 2 * 3 * 5 * 7 * 13 * 19 * 37 37 69090840
38 6 2 * 3 39 228
40 13530 2 * 3 * 5 * 11 * 41 41 541200
42 1806 2 * 3 * 7 * 43 43 75852
44 690 2 * 3 * 5 * 23 45 30360
46 282 2 * 3 * 47 47 12972
48 46410 2 * 3 * 5 * 7 * 13 * 17 49 2227680
50 66 2 * 3 * 11 51 3300
52 1590 2 * 3 * 5 * 53 53 82680
54 798 2 * 3 * 7 * 19 55 43092
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56 870 2 * 3 * 5 * 29 57 4872058 354 2 * 3 * 59 59 20532
6056786730
2 * 3 * 5 * 7 * 11 * 13 * 31
* 61 613407203800
62 6 2 * 3 63 372
64 510 2 * 3 * 5 * 17 65 32640
66 64722 2 * 3 * 7 * 23 * 67 67 4271652
68 30 2 * 3 * 5 69 2040
70 4686 2 * 3 * 11 * 71 71 328020
72140100870
2 * 3 * 5 * 7 * 13 * 19 * 37* 73 73
10087262640
74 6 2 * 3 75 444
76 30 2 * 3 * 5 77 2280
78 3318 2 * 3 * 7 * 79 79 258804
80 230010 2 * 3 * 5 * 11 * 17 * 41 81 18400800
82 498 2 * 3 * 83 83 40836
84 3404310 2 * 3 * 5 * 7 * 13 * 29 * 43 85 285962040
86 6 2 * 3 87 516
88 61410 2 * 3 * 5 * 23 * 89 89 540408090 272118 2 * 3 * 7 * 11 * 19 * 31 91 24490620
92 1410 2 * 3 * 5 * 47 93 129720
94 6 2 * 3 95 564
96 4501770 2 * 3 * 5 * 7 * 13 * 17 * 97 97 432169920
98 6 2 * 3 99 588
100 33330 2 * 3 * 5 * 11 * 101 101 3333000
498 3499986 2 * 3 * 7 * 167 * 499 499 1742993028
We note that all the numbers of the column den (ς(1-n)) in bleu are divisible for 24, i.e.the number that is related to the physical vibrations of the bosonic strings by the
following Ramanujan function:
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( )
++
+
⋅
=
−
∞ −
∫
4
2710
4
21110log
'
142
'
cosh
'cos
log4
24
2
'
'4
0
'
2
2
wt itwe
dxe x
txw
anti
w
wt
w x
φ
π π
π
π
.
So we have a relation between the den(Bn) and the prime numbers that are hidden in the
den(Bn).
From the recursive formula
Bm = 1-∑−
=
1
0
m
k k
m Bk /(m-k+1)
For example:
B1 = 1-B0 /2=1-1/2=1/2
B2 = 1-B0 /3-(2/2)B1=1-1/3-1/2=1/6
B3 = 0
B4 = 1-B0 /5-(4/4)B1-(6/3)B2= 1- 1/5-1/2-1/3 = -1/30
B5 = 0
B6 = 1-B0 /7-(6/6)B1-(15/5)B2-(15/3)B4=1-1/7-1/2-1/2+1/6=1/42
B7 = 0
B8 = 1-B0 /9-(8/8)B1-(28/7)B2-(70/5)B4-(28/3)B6 = 1-1/9-1/2-4/6+14/30-2/9 = -3/90 = -
1/30
B4 = B8 = -1/30
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We can find any Bm starting with all the previous Bm-1.
It follows that any prime number can be found by knowing all the prime numbers thatprecede it.
We can predict the next prime number knowing all the prime numbers that precede it.
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1.1.1 CALCULATION OF den(B N )
1° method:
The denominator den( Bn) is thus the product of all divisors di+1 of the index of Bn
minus those who aren't prime number, except 2
In fact we have:
Example:
n=6
divisors of 6 di: 1, 2, 3, 6
prime numbers di+1: 2, 3, 4, 7
4 is discarded, because is not a prime number, and so:
den( B6 ) = 2*3*7
2° method:
Another way to calculate the den( Bn) considering all the product of prime numbers up ton+1 and remove the ones that are not the divisors di-1 of the index.
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den( B6 ) = 2*3*5*7 (all the prime up to n+1)
divisors di-1: 1, 2, 4, 6
4 is discarded because it isn't a divisor of 6 and so we discard 5 in the product:
den( B6 ) = 2*3*7
We can write:
den( Bn) = p∏ with p-1 divisors of n
Observation:
In both two methods in den(Bn) are always the factors 2 and 3 because all den(Bn) aremultiple of 6.
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Example:
For calculate den( B498) with the first method:
n=498
divisors of 498 di: 1, 2, 3, 6, 83, 166, 249, 498
prime numbers di+1: 2, 3, 4, 7, 84, 167, 250, 499
4, 84 and 250 are discarded, because not prime numbers, and so:
den( B498) = 2*3*7*167*499 = 3.499.986
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1.1.2 NUMERATOR OF B N num(B N )
We can define:
Numerator( B2n) = num( B2n) → n
num( B 2n ) = first factor of the factorization( B2n ) if the first number is n we have that
it is a prime number p
Example:
n=11
num(B22)= 854.513
Numerator( B22)= first factor of the factorization( B22) = 854.513 = 11*131*593 → 11prime number.
When n odd integers we have that all the values of Bn are zero and so we find all
the n+1 even integer.
When we have a num( B 2n ) that gives a number n that isn’t a prime number or
multiple of only 2 and 3, the factorization of num( B 2n ) is starting with a new prime
number.
If n=kp or a multiple of a prime number, but not multiple of only 2 and 3, we have
that the factorization of num( B 2n ) is always starting with p
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Example:
n=14
num(B28)= 23.749.461.029
Numerator( B28)= first factor of the factorization( B28) = 23.749.461.029=
7*9349*362903 → 7 prime number
Index of B28 → 28=4*7, 7 prime number
----------------------------------------------------------------------------------------------------------
n=16
num(B32)= 7.709.321.041.217
Numerator( B32)= first factor of the factorization( B32) = 7.709.321.041.217=
37*683*305065927 → 16 is not valid
Index of B32 → 32=25, not valid because it isn’t a multiple of a prime number, it is
multiple of only 2.
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TAB. 2
2n num(B2n) Factorization prime frequency
10 5 5 5 20k+10
12 691 691 7 42k+14, 42k+28
14 7 7 11 22k
16 3617 3617 13 26k
18 43867 43867 17 34k
20 174611 283 617 19 38k
22 854513 11 131 593 23 46k
24 236364091 103 2294797 29 58k
26 8553103 13 657931 31 62k
28 23749461029 7 9349 362903 37 36k+32, 74k
30 8,61584E+12 5 1721 1001259881 41 82k
32 7,70932E+12 37 683 305065927 43 86k
34 2,57769E+12 17 151628697551 47 94k
36 2,63153E+19 2,63153E+19 53 106k
38 2,92999E+15 19 154210205991661 59 59k+44, 118k
40 2,61083E+20 137616929 1897170067619 61 122k
42 1,5201E+21 1,5201E+21 67 66k+58, 134k
44 2,78333E+22 11 59 8089 2947939 1798482437 71 142k
46 5,96451E+23 23 383799511 67568238839737 73 146k
48 5,6094E+27 653 56039 153289748932447906241 79 158k
50 4,95057E+26 5^2 417202699 47464429777438199 83 166k
52 8,01166E+2913 577 58741 4010291774534045619429
89 178k
54 2,915E+31 39409 97 194k
56 2,47939E+337 113161 163979
19088082706840550550313
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58 8,44836E+3429 67 186707 623524204937349583369104129
60 1,21523E+42 2003
62 1,23006E+37 31 157 266689
64 1,06784E+41 1,06784E+41
66 1,4726E+45
11 839
1595622518286201813903585901562
39282938769
68 7,87731E+4317 37 101 123143 1822329343
5525473366510930028227481
70 1,50538E+48 5 7 688531
72 5,82795E+54 5,82795E+54
74 3,41524E+49
37
9230383051140856220089209116614
22572613197507651
76 2,46551E+52 19 58231
78 4,14846E+56 1380 4,60378E+60 631 10589
We can observe that irregular primes have a different behavior from other regular
primes in their frequencies.
In fact irregular primes 37, 59, 67,… have two different frequencies while all other
regular primes have only one regular frequency:
for p prime number we have frequency = 2p
We have an exception with the number prime 7. It has two frequencies but both are
regular. In fact 7 occurs with a frequency of:
2n = 14, 28, 56, 70, 98, 112, …or 42k+14, 42k+28
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1.1.3 FINDING A PROBABLY PRIME NUMBER p LARGE AS DESIRED TO
60%
We can find a large prime number as we like by simply considering this formula withthe den( Bn)
p = den( Bn)+1 or p = den( Bn)-1
We thus have two prime numbers or only one that is a prime number, or none of the twois a prime number.
The latter case is, however, the least likely of the 3 possible cases.
In the following table are highlighted the prime numbers that derive:
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TAB. 3
n den(Bn) Factorization den(Bn)-1 den(Bn)+1
1 2 2 1 3
2 6 2 * 3 5 7
4 30 2 * 3 * 5 29 31
6 42 2 * 3 * 7 41 43
8 30 2 * 3 * 5 29 3110 66 2 * 3 * 11 65 67
12 2730 2 * 3 * 5 * 7 * 13 2729 2731
14 6 2 * 3 5 7
16 510 2 * 3 * 5 * 17 509 511
18 798 2 * 3 * 7 * 19 797 799
20 330 2 * 3 * 5 * 11 329 331
22 138 2 * 3 * 23 137 139
24 2730 2 * 3 * 5 * 7 * 13 2729 2731
26 6 2 * 3 5 728 870 2 * 3 * 5 * 29 869 871
30 14322 2 * 3 * 7 * 11 * 31 14321 14323
32 510 2 * 3 * 5 * 17 509 511
34 6 2 * 3 5 7
36 19191902 * 3 * 5 * 7 * 13 * 19 *
37 1919189 1919191
38 6 2 * 3 5 7
40 13530 2 * 3 * 5 * 11 * 41 13529 13531
42 1806 2 * 3 * 7 * 43 1805 180744 690 2 * 3 * 5 * 23 689 691
46 282 2 * 3 * 47 281 283
48 46410 2 * 3 * 5 * 7 * 13 * 17 46409 46411
50 66 2 * 3 * 11 65 67
52 1590 2 * 3 * 5 * 53 1589 1591
54 798 2 * 3 * 7 * 19 797 799
56 870 2 * 3 * 5 * 29 869 871
58 354 2 * 3 * 59 353 355
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6056786730
2 * 3 * 5 * 7 * 11 * 13 *31 * 61 56786729 56786731
62 6 2 * 3 5 7
64 510 2 * 3 * 5 * 17 509 511
66 64722 2 * 3 * 7 * 23 * 67 64721 64723
68 30 2 * 3 * 5 29 31
70 4686 2 * 3 * 11 * 71 4685 4687
721,4E+08
2 * 3 * 5 * 7 * 13 * 19 *
37 * 73 140100869 140100871
74 6 2 * 3 5 776 30 2 * 3 * 5 29 31
78 3318 2 * 3 * 7 * 79 3317 3319
80 230010 2 * 3 * 5 * 11 * 17 * 41 230009 230011
82 498 2 * 3 * 83 497 499
843404310
2 * 3 * 5 * 7 * 13 * 29 *43 3404309 3404311
86 6 2 * 3 5 7
88 61410 2 * 3 * 5 * 23 * 89 61409 61411
90 272118 2 * 3 * 7 * 11 * 19 * 31 272117 27211992 1410 2 * 3 * 5 * 47 1409 1411
94 6 2 * 3 5 7
964501770
2 * 3 * 5 * 7 * 13 * 17 *
97 4501769 4501771
98 6 2 * 3 5 7
100 33330 2 * 3 * 5 * 11 * 101 33329 33331
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1.2 CONNECTION ζ (s) ZETA RIEMANN FUNCTION WITH PRIME NUMBERS
Bernoulli numbers as values of the Riemann zeta function.
Associated sequence: 1, +1/2, 1/6, 0, …
Using this convention, the values of the Riemann zeta function satisfy
ζ(1−n)= -n
Bn
for all integers n≥0.
For n=0 is to be understood as B0=lim x→0 xζ(1− x)=1
So we have:
Denominator(nζ(1−n)) = den(nζ(1−n)) = den( Bn) → n+1
We can eliminate the minus sign “-” because we consider only the positive integers asdenominator.
Denominator (nζ(1−n))= last factor of the factorization (nζ(1−n)) if the last number is n
+1 we have that it is a prime number p;
or also:
den(ζ(1−n))= last factor of the factorization (ζ(1−n)) if the last number is n +1 we havethat it is a prime number p
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Obviously, for negative even integers, we have that:
ζ(−2n)= 0 for n ≥ 1
ζ vanishes at the negative even integers because Bm = 0 for all odd m other than 1
So there is a link with
ζ(−2n−1) for n ≥ 0
or with the denominator ζ(negative odd integers) and the prime numbers:
ζ(0) = -2
1→ det(2) → 2
ζ(−1) = -12
1→ det(12) = 2
2 * 3 → 3
ζ(−3) = +120
1→ det(120) = 2
3 * 3 * 5 → 5
ζ(−5) = -252
1→ det(252) = 2
2 * 3
2 * 7 → 7
Being present n we have a factorization with exponents while with regard to Bn wereonly distinct prime factors with exponent equal to 1.
We note that for
ζ(−3) = +120
1→ det(120) = 2
3 * 3 * 5 → 5
we have that 120 = 24 * 5 , where 24 is related to the physical vibrations of the bosonic
strings by the following Ramanujan function:
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( )
++
+
⋅
=
−
∞−
∫
4
2710
4
21110log
'
142
'
cosh
'cos
log4
24
2
'
'4
0
'
2
2
wt itwe
dxe x
txw
anti
w
wt
w x
φ
π
π
π
π
,
and that 2, 3 and 5 are all Fibonacci’s numbers.
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1.3.1 SPECIFIC VALUES
Case s=-1
ζ(-1) = 1+2+3+4+…..= ...6
7
4
5
2
32 =
s prime p
−−
∏1
1= -
12
1
gives a way to assign a finite result to the divergent series 1 + 2 + 3 + 4 + …..·
----------------------------------------------------------------------------------------------------------
Case s=0
ζ(0) = 1+1+1+1+…. =0
1
0
1
0
1
0
1…. = -
2
1
gives a way to assign a finite result to the divergent series 1+1+1+1+….
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Case s=1, Pole of ζ(s)
( ) ∞=+++++= ...5
1
4
1
3
1
2
111ζ
thus,
∞=⋅⋅⋅⋅⋅
⋅⋅⋅⋅⋅
=+++++ ...106421
...117532
...5
1
4
1
3
1
2
1
1
for x→1+ ζ(1
+)=+∞
for x→1- ζ(1
-)=-∞
B0=lim x→0 xζ(1− x)=1
----------------------------------------------------------------------------------------------------------
Case s=2
( ) 645.16
...3
1
2
112
2
22 ≈=+++=
π ζ ;
ζ(2) = 48
49
24
25
8
9
3
4
…. = 6
2π
We note that 1.645 is very near to the aurea frequency 1.64809064 that is given by
0,61803398 · 4 · 2/3, where 0,61803398 is the aurea section, i.e.2
15 −
----------------------------------------------------------------------------------------------------------
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Case s=+∞
ζ(+∞) = 1+
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2. BERNOULLI'S FORMULA EXTENDED IN THE FIELD OF REAL
NUMBERS x ∈R
We can define the Bernoulli's formula as the analytic continuation of the formula in thefield of real numbers ∈ x R, as it had done with the Riemann zeta function.
The Cartesian plane of the zeta function ζ(x) for real numbers ∈R between -18.5≤ n ≤10
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We know that
ζ(1−n)= -n
Bn
for all integers n≥0.
We extend this formula with ∈ x R considering also the negative values of Bernoulli’snumbers and all the real positive and negative numbers, and so we have:
x B = ( ) x B = x− ζ ( ) x−1
We must consider the zeta function ζ(1-x), shifted to the left side of one, so that the y-axis becomes the x=0 pole.
Then we must draw the graph symmetric to the y-axis as if it were reflected.
Finally, we must multiply by the function bisector line of the second and fourthquadrant y=-x
The table with -10≤x≤25 is the following:
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TAB. 5
x B(x)
-10 10,004941
-9 9,008951
-8 8,016000
-7 7,028000-6 6,050094
-5 5,071262
-4 4,147600
-3 3,246969
-2 2,404113
-1 1,644934
0 1
1 0,500000
2 0,1666673 0
4 -0,033333
5 0
6 0,023810
7 0
8 -0,033333
9 0
10 0,075758
11 012 -0,253114
13 0
14 1,166667
15 0
16 -7,092156863
17 0
18 54,971177945
19 0
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20 -529,124242421 0
22 6192,123188
23 0
24 -86580,25311
25 0
And the graph is:
In x=0 is the pole.
We note that for x→-∞ ( ) x x B x −=−∞→lim
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and for x→+∞ we have oscillations increasingly large and always larger with all oddvalues of B(x) equal to 0.
For negative integers values:
B(-1/2)= 1/2 (3/2)= (1/2)*2,612...= 1,306
B(-1)= ζ(2)= 6
2π
= 1,6449...
B(-3/2)= 3/2 (5/2)= (3/2)*1,341...= 2,01...
B(-2)= 2ζ(3)= 2* 1,202056…= 2,404113…
B(-5/2) = 5/2 (7/2)= (5/2)*1,127= 2,8175...
B(-3)= 3ζ(4)= 3*
90
4π =3,246969
This value is very near to the following aurea frequency 3,24611797 that is equal about
to (1,61803398…)-5
· 12 · 3 = 3,246118, where2
15...61803398,1
+= is the aurea ratio.
…
B(-x)≈ -x per x large x>10
For positive integers values:
B(0)= 0*ζ(1)= 1= lim x→0 xζ(1− x)
B(1/2)= -1/2 (1/2)=-(1/2)(-1.4603...)= 0,73..
B(1)= -ζ(0)= ½
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B(2)= -2ζ(-1)= 1/6
B(3)= -3ζ(-2)= 0
B(4)= -4 (-3)= -1/30
So all the odd integers of B(x)=0 and therefore coincide with the even negative
integers trivial zeros of ζ(1-x)=0
for x=2k+1
k = 1, 2, 3, ...
B(2k+1)=ζ(-2k)=0
or
x=3, 5, 7, ....
B(x)=ζ(1-x)=0
The (1-x) even negative integers of ζ(1-x), with x=3, 5, 7, ...., match the negative
axis y=0 of the Cartesian plane and coincide with the zeros of the odd integer of
B(x).
In the range 0≤x≤12 is the following
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3. BERNOULLI's FORMULA EXTENDED IN THE FIELD OF COMPLEX
NUMBERS z∈C, z=σ ±it
We can define the Bernoulli's formula as the analytic continuation of the formula in thefield of complex numbers z∈C, z=σ ±it , as it had done with the Riemann zeta function.
We extend this formula with z∈C, z=σ ±it , considering also complex values ofBernoulli’s numbers and so we have:
B z = B(z)=-zζ(1− z)
( ) ∑=
+++==n
k
mmmm
m nk nS 1
...21 .
for m, n ≥ 0
We know that the Riemann zeta function ζ (s) is a function of a complex variable s=σ +it .
The following infinite series converges for all the complex numbers s with real partgreater than 1, or Re(s)>1, and defines ζ (s) in this case:
( ) ∑∞
=
−+++==
1
...31
21
11
nsss
snsζ ( ) 1>= sR σ .
We have
ζ(1−n)= -n
Bn
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for all integers n>0.
If we compare the 2 function Sm(n) for n→+∞ and ζ(s) the single factors of the twoseries is one the reciprocal of the other, or equivalently
ζ(-n) =∑∞
=1k
nk
for all integers n≤0.
for all integers n≤0.
Besides we have for s>1:
for s<1 we can write in the same way:
ζ(-s) = ...71
1
51
1
31
1
21
1ssss
−−−−=
s prime p−
∏1
1
Going from the product representation of the Riemann zeta-function and the aboveseries expansions of the Bernoulli numbers, the following shows:
( )
( )
( )
( )∏∈
−
−
−
−
=
−=
P p
nnn
nnnn
n
p
n
...5
11
3
11
2
11
1
2
!2211
2
!22
222
2
1
22π π
β
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for n≥0
Here, the product extends over all primes (Euler product of the Riemann zeta function)and
βn = |B(2n)| the absolute value
We can calculate, for example, B(2)
n=1
B(2) = β1 =( )
2
2 11
1
)2(
!22
p
prime
−
∏π
=64
4 2
2
π
π =
6
1
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4. ZEROS ζ(½ ± ix)=0 ON THE CRITICAL LINE
The functional equation shows that the Riemann zeta function has zeros at −2, −4, ... .
These are called the trivial zeros. They are trivial in the sense that their existence is
relatively easy to prove, for example, from sin(πs /2) being 0 in the functional equation
(see below). The non-trivial zeros have captured far more attention because theirdistribution not only is far less understood but, more importantly, their study yieldsimpressive results concerning prime numbers and related objects in number theory.
The Riemann zeta function satisfies the functional equation:
( ) ( ) ( )sss
sss
−−Γ
=
− 112
sin2 1 ζ π
π ζ ,
where Γ(s) is the gamma function, which is an equality of meromorphic functions validon the whole complex plane.
It is known that any non-trivial zero lies in the open strip {s∈C: 0<Re(s)<1}, which is
called the critical strip. The Riemann hypothesis, asserts that any non-trivial zero s has
Re(s)=1/2. In the theory of the Riemann zeta function, the set {s∈C: Re(s)= 1/2} is
called the critical line.
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Blue is the real part and red is the imaginary part of the function ζ(½ ± ix) for 0≤x≤100
is shown so that we can clearly see the first non-trivial zeros and the values where real
and imaginary part are both equal to zero:
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Re(ζ(½±ix)) = IM(ζ(½±ix)) = 0
We have for all ∈s C\ { }1 the integral relation
( ) ( )
( ) ( )∫∞
−
++
−
−
=0
22
1
11
arctansin2
1
2dt
et
t s
s
s
t
s
ss
π
ζ ,
holds true, which may be used for a numerical evaluation of the zeta-function.
To find non-trivial zeros for ς(p) = 0 and substituting s with p (s=p) we must have
p = x + jy
ς(x + jy) =yj1-x
2 yj1-x
+
+
- yjx2 +
∫∞
+++
+
0
2 / 2 / 2 )1()1(
)arctan*)sin((dt
et
t yj xt jy x π
dividing by 2p ≠ 0 (always true) we have
1)-2(p1 = ∫
∞
++0
2 / 2 )1()1()arctansin( dt
et
t pt p π
Only with particular values of x=½ and y that are precisely the zeros of the function we
have:
for p= ½ + jy, with y that has a specific real value ∈ R
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∫∞
++0
2 / 2 )1()1(
)arctansin(dt
et
t pt p π
= ∫∞
−
−
0
2)1(2
1dt
p=
1)-2(p
1;
2yj1-
1
+= ∫
∞
+++
+
0
2 / y4 / 12 )1()1(
)arctan)y2
1sin((
dt et
t j
t j π =
= )
4
12
4
1y
4
12
(2
2
2
0
2
2
+
−
+
+
∫∞
y
j
y
ydy=
+ 4
1
4
1-
2
y
-j
+ 4
1
2
y
2
y
=24y1
1-
+ - j
24y1
2y
+
for the first zero of the Riemann zeta function:
y=14,134725...
p= ½ + j14,134725...
2yj1-
1
+ = .j28,26945..1-
1
+ = .800,1618..
1-
- ...1618,800
.y28,26945..
j = -0.0012497 - 0.0353296j
∫∞
+++
+
0
2 / 14,1347254 / 12 )1()1(
)arctan)14,1347252
1sin((
dt et
t j
t j π =
24y1
1-
+ - j
24y1
2y
+= -0.0012497 - 0.0353296j
The two quantities are equal and p = ½+j14.134725... is a non-trivial zero, and is also
the first zero of the Riemann zeta function.
In the table the first 30 zeros of Riemann zeta function ς(½ + jy) with y up to 100
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TAB. 6
n°Value of the IM
of zero, y
1 14,134725142
2 21,022039639
3 25,010857580
4 30,424876126
5 32,935061588
6 37,586178159
7 40,918719012
8 43,327073281
9 48,005150881
10 49,773832478
11 52,970321478
12 56,446247697
13 59,347044003
14 60,831778525
15 65,112544048
16 67,079810529
17 69,546401711
18 72,067157674
19 75,704690699
20 77,144840069
21 79,33737502022 82,910380854
23 84,735492981
24 87,425274613
25 88,809111208
26 92,491899271
27 94,651344041
28 95,870634228
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29 98,83119421830 101,317851006
Returning to the Bernoulli numbers we have:
B z = B(z)=-zζ(1− z)
with z=½ ± ix
B(½ + ix)=-(½ + ix)ζ(½ - ix)
B(½ - ix)=-(½ - ix)ζ(½ + ix)
Let's consider the zeros of ζ(½ + ix) with the particular values of x=b so that ζ(½+ib)=0.
So all the values of z=½ + ib of ζ(½+ib)=0 and therefore coincide with the values of
z=½ + ib of B(½-ib)=0 or with its complex conjugate.
The complex zeros of ζ(½+ib)=0 coincide with the complex conjugate zeros of B(½-ib)=0 and vice versa.
Example for the first three zeros
b=14,134725142 – 21,022039639 - 25,010857580.....
ζ(½+ib)=B(½-ib)=0
ζ(½-ib)=B(½+ib)=0
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But the function ζ(½ + ix) has others values where the IM(ζ(½ + ix))=0 and so we findthe correlation with the Gram points g IM(ζ(½ + ig))=0
The first Gram point is:
B(½ - i3,436)=-(½ - i3,436)ζ(½ + i3,436)
or
B(½ + i3,436)=-(½ + i3,436)ζ(½ - i3,436)
ζ(½ + i0) = -2B(½) ≈-1,4603545...
ζ(½ + i3,436)≈0,5641
ζ(½ + i9,667)≈1,53181
ζ(½ + i14,13)=0 the first zero of ζ(z)=0
ζ(½ + i17,8455) ≈ 2,34018
ζ(½ + i21,02)=0 the second zero of ζ(z)=0
ζ(½ + i23,17)≈1,45744
ζ(½ + i25,01)=0 the third zero of ζ(z)=0
ζ(½ + i27,67)≈2,84509
These are the Gram points.
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TAB. 7 with Gram Points:
x jy
0,000000 -1,46035
3,436218 -0,56415
9,666908 -1,53181
14,134725 017,845599 2,34018
21,022040 0
23,170283 -1,45744
25,010858 0
27,670182 2,84509
30,424876 0
31,717980 -0,92526
32,935062 0
35,467184 2,9381237,586178 0
38,999210 -1,78672
40,918719 0
42,363550 1,098756
43,327073 0
45,593029 -3,6629
48,005151 0
48,710777 0,688292
49,773832 051,733843 -2,01121
52,970321 0
54,675237 2,91239
56,446248 0
57,545165 -1,75816
59,347044 0
60,351812 0,53858
60,831779 0
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63,101868 -4,164465,112544 0
65,800888 1,053877
67,079811 0
68,453545 -1,54
69,546402 0
71,000000 1,9
72,067158 0
73,600000 -3,6
75,704691 076,200000 0,7
77,144840 0
78,700000 -1,3
79,337375 0
81,100000 3,97
82,910381 0
83,500000 -1,2
84,735493 0
86,000000 1,0987,425275 0
88,300000 0,8
88,809111 0
90,800000 3,97
92,491899 0
93,100000 1,39
94,651344 0
95,300000 0,7
95,870634 097,700000 2,9
98,831194 0
100,000000 2,8
101,317851 0
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4.1 GRAPHICS OF THE REAL AND IMAGINARY ς (p)
Re ς (1/2 + jy) and Im ς (1/2 + jy)
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Re ς (1/3 + jy) and Im ς (1/3 + jy)
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Re ς (3/4 + jy) and Im ς (3/4 + jy)
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5. A NEW REPRESENTATION OF THE BERNOULLI FUNCTION
CONNECTED WITH THE RIEMANN ZETA FUNCTION
Applying for all ∈s C\ { }1 the integral relation
( ) ( )
( ) ( )∫∞
−
++
−−
=0
22
1
11
arctansin2
1
2dt
et
t s
ss
t s
ss
π
ζ ,
and
B(z)=-zζ(1− z)
substituting s with z (z=p) we have
1-z → z
ς(1-z) = z
z 2*
1
− -
z2
2∫∞
−
++
−
0 2
1
2 )1()1(
)arctan*)1sin((dt
et
t z
t
z
π
B(z)= z21 +
z z
22 ∫
∞
−
++
−
0 2
1
2 )1()1(
)arctan*)1sin(( dt
et
t z
t
z
π
or
B(z)= z2
1[1 + 2z ∫
∞
−
++
−
0 2
1
2 )1()1(
)arctan*)1sin((dt
et
t z
t
z
π
]
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B(z)=-zζ(1− z)
z)-(1
B(z)
ζ =-z
ζ(½+ib)=B(½-ib)=0
ζ(½-ib)=B(½+ib)=0
To find non-trivial zeros for B(z)= 0
with
z = x + jy
2z ≠ 0 (always) we have
[1 + 2z ∫∞
−
++
−
0 2
1
2 )1()1(
)arctan*)1sin((dt
et
t z
t
z
π
] = 0; z2
1− = ∫
∞
−
++
−
0 2
1
2 )1()1(
)arctan*)1sin((dt
et
t z
t
z
π
That’s equivalence with (pag. 40)
1)-2(p1 = ∫
∞
++0
2 / 2 )1()1()arctansin( dt
et t p t p π
with 1-p →z
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Now, we have:
∫∞
++0
2 / 2 )1()1(
)arctansin(dt
et
t pt p π
= ∫∞
−
−
0
2)1(2
1dt
p=
1)-2(p
1;
For p = 1/2+yj:
∫∞
+++
+
0
2 / y4 / 12 )1()1()arctan)y2
1
sin(( dt et
t jt j π
=2yj1-
1+
=24y1
1-+
- j24y1
2y+
Multiplying this equation for 1/6, we obtain the following interesting expression:
6
1∫∞
+++
+
0
2 / y4 / 12 )1()1(
)arctan)y2
1sin((
dt et
t j
t j π =
24y26
1-
+ - j
24y26
2y
+;
This expression can be related with the number 24 that is connected to the “modes” thatcorrespond to the physical vibrations of the bosonic strings by the following Ramanujan
function:
( )
++
+
⋅
=
−
∞−
∫
4
2710
4
21110log
'
142
'
cosh
'cos
log4
24
2
'
'4
0
'
2
2
wt itwe
dxe x
txw
anti
w
wt
w x
φ
π
π
π
π
.
Thence, we have the following mathematical connection:
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6
1∫∞
+++
+
0
2 / y4 / 12 )1()1(
)arctan)y21sin((
dt et
t j
t j π =
24y26
1-
+ - j ⇒
+24y26
2y
( )
++
+
⋅
⇒
−
∞−
∫
4
2710
4
21110log
'
142
'
cosh
'cos
log42
'
'4
0
'
2
2
wt itwe
dxe x
txw
anti
w
wt
w x
φ
π
π
π
π
.
5.1 A REPRESENTATION OF THE ABSOLUTE VALUES OF BERNOULLI
NUMBERS CONNECTED WITH THE RIEMANN ZETA FUNCTION FOR n≥ 1
The *
n B =|B(2n)| absolute values of Bernoulli numbers may be calculated from the
integral:
∫∞
−∗
−=
0 2
12
14
t
n
ne
dt t n B
π ,
for n≥1*
n B = |B(2n)| the absolute value
from
B(z)=-zζ(1− z)
with 2n → z, we have
|B(2n)|= 2n|ζ(1−2n)|
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|ζ(1−2n)| = 2 ∫∞ −
−0
2
12
1dt
e
t t
n
π for n≥1
Multiplying this equation for 4, we obtain the following interesting relationship:
4 |ζ(1−2n)| = 8 ∫∞ −
−0
2
12
1dt
e
t t
n
π
We remember that 8 is a Fibonacci’s number and is connected with the “modes” that
correspond to the physical vibrations of a superstring by the following Ramanujan
function:
( )
++
+
⋅
=
−
∞−
∫
4
2710
4
21110log
'
142
'
cosh
'cos
log4
318
2
'
'4
0
'
2
2
wt itwe
dxe x
txw
anti
w
wt
w x
φ
π
π
π
π
,
We have the following new equation:
4 |ζ(1−2n)| = ( )
++
+
⋅
−
∞−
∫
4
2710
4
21110log
'
142
'
cosh
'cos
log4
31
2
'
'4
0
'
2
2
wt
itwe
dxe x
txw
anti
w
wt
w x
φ
π
π
π
π
∫∞ −
−0
2
12
1dt
et
t
n
π
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6. REFERENCES
1) Wikipedia
2) Tesi di laurea su i numeri Bernoulli e loro applicazioni
3) PROGETTO BEFZS POSSIBILI CONNESSIONI MATEMATICHE TRA:
a) i numeri Bernoulli (B), b) i numeri di Eulero (E),
c) i numeri di Fibonacci (F),
d) la funzione zeta (Z) di Riemann, e
e) la teoria di stringa (S)