1
On some equations concerning some topics of Field Theory and Gravity and
applications of κ formula regarding the Zeros of Riemann Zeta Function.
Mathematical connections with some sectors of String Theory and Ramanujan
Mathematics.
Michele Nardelli
1
, Antonio Nardelli
2
Abstract
In this research thesis, we analyze some topics of Field Theory and Gravity and
applications of κ formula regarding the Zeros of Riemann Zeta Function. We
describe the mathematical connections with some sectors of String Theory and
Ramanujan Mathematics.
1
M.Nardelli studied at Dipartimento di Scienze della Terra Università degli Studi di Napoli Federico II,
Largo S. Marcellino, 10 - 80138 Napoli, Dipartimento di Matematica ed Applicazioni “R. Caccioppoli” -
Università degli Studi di Napoli “Federico II” – Polo delle Scienze e delle Tecnologie Monte S. Angelo, Via
Cintia (Fuorigrotta), 80126 Napoli, Italy
2
A. Nardelli studies at the Università degli Studi di Napoli Federico II - Dipartimento di Studi Umanistici –
Sezione Filosofia - scholar of Theoretical Philosophy
2
Anatoly A. Karatsuba
(1937-2008)
(Mathematician)
Vesuvius landscape with gorse – Naples
4
5
From:
On the Zeros of the Davenport Heilbronn Function
S. A. Gritsenko - Received May 15, 2016 - ISSN 0081-5438, Proceedings of the
Steklov Institute of Mathematics, 2017, Vol. 296, pp. 65–87.
We have:
(
10 2
5 2) (
5 1 ) =
6
Input:
Decimal approximation:
0.28407904384…. = κ
Alternate forms:
Minimal polynomial:
Expanded forms:
7
For ((((√(10-2√5) -2))⁄((√5-1)))) = 8πG; G = 0.011303146014
Indeed:
((((√(10-2√5) -2))⁄((√5-1))))/(8π)
Input:
Result:
Decimal approximation:
0.01130314…. = g (gravitational coupling constant)
Property:
8
Alternate forms:
Expanded forms:
Series representations:
9
We note that:
(((√(10-2√5) -2))⁄((√5-1)))*((2 i (sqrt(5) - 1) t + sqrt(5) - 1)/(2 (sqrt(2 (5 - sqrt(5))) -
2)))
Input:
Exact result:
10
Plot:
Alternate form assuming t>0:
Alternate forms:
11
1/2+it = real part of every nontrivial zero of the Riemann zeta function
Derivative:
Indefinite integral:
And again:
(((√(10-2√5) -2))⁄((2x)))*((2 i (sqrt(5) - 1) t + sqrt(5) - 1)/(2 (sqrt(2 (5 - sqrt(5))) - 2)))
= (1/2+it)
Input:
Exact result:
12
Alternate form assuming t and x are real:
Alternate form:
Alternate form assuming t and x are positive:
Expanded forms:
Solutions:
13
Input:
Decimal approximation:
0.6180339887…. =
1
Solution for the variable x:
Implicit derivatives:
14
From:
Course of Field Theory and Gravity - Prof. Augusto Sagnotti (SNS Pisa-Italy)
We have that:
15
2
2/2
4
/2
∞
0
2
/2
=
2
2/2
2
2
1
4
/2
/2
= Γ
1 /2
∞
0
=
2
2
2/2
2
2
1
Γ
1/2
4
/2
=
2
2
2
12
Γ
1+
4
2
2
2
32
2
From:
2
2/2
2
2
1
4
/2
/2
= Γ
1 /2
∞
0
For D = 10:
λ(μ^2 )^(2-10/2) (0.511^2 )^(10/2-1)/(4π)^(10/2) integrate((e^(-s) s^(-10/2))) =
gamma(-1+10/2)
Input:
Result:
16
(6.18168×10^-10 e^(-1/2) (e^(1/2) (1/2)^4 Ei(-1/2) + (1/2)^3 - (1/2)^2 + 2 (1/2) -
6))/((1/2)^4 λ(μ^2)^3) = 6
Input interpretation:
Result:
-(3.10909×10^-8)/x(y^2)^3 = 6
Input interpretation:
Result:
17
Implicit plot:
Alternate form:
Alternate form assuming x and y are real:
Solutions:
18
Integer solutions:
Solutions for the variable y:
Implicit derivatives:
19
from
And
-(3.10909×10^-8)/(((-5.18182×10^-9 y^6 )(y^2)^3)) = 6
Input interpretation:
Result:
Alternate form assuming y is positive:
Alternate form assuming y is real:
Real solutions:
20
Complex solutions:
-(3.10909×10^-8)/(((-5.18182×10^-9 )))
Input interpretation:
Result:
Repeating decimal:
5.99999614035…. ≈ 6
21
With D = 11:
λ(μ^2 )^(2-11/2) (0.511^2 )^(11/2-1)/(4π)^(11/2) integrate((e^(-s) s^(-11/2))) =
gamma(-1+11/2)
Input:
Result:
where
is equal to
(105 sqrt(π))/16
Input:
22
Exact result:
Decimal approximation:
11.631728…
Property:
Series representations:
For s =1/2 , we obtain:
(4.52618×10^-12 e^(-1/2) (16 e^(1/2)*(1/2)^(9/2) Γ(1/2, 1/2) - 16 (1/2)^4 + 8 (1/2)^3
- 12 (1/2)^2 + 30 *1/2 - 105))/((1/2)^(9/2) λ(μ^2)^(7/2)) = (105 sqrt(π))/16
23
Input interpretation:
Result:
-(5.73627×10^-9)/((x(y^2)^(7/2))) = (105 sqrt(π))/16
Input interpretation:
Result:
24
Implicit plot:
Alternate form assuming x and y are real:
Alternate form:
Alternate form assuming x and y are positive:
25
Expanded forms:
Real solutions:
Solutions for the variable y:
26
27
Implicit derivatives:
From:
For x = -1 and
-(5.73627×10^-9)/((-(((0.0291915 + 0.0366049 i)/(-1)^(1/7))^2)^(7/2)))
Input interpretation:
Result:
Polar coordinates:
11.6317
28
Polar forms:
For D = 7 :
λ(μ^2 )^(2-7/2) (0.511^2 )^(7/2-1)/(4π)^(7/2) integrate((e^(-7) s^(-7/2))) = gamma(-
1+7/2)
Input:
Result:
-(1.80663×10^-9)/(0.5^(5/2) λ(μ^2)^(3/2)) = (3 sqrt(π))/4
Input interpretation:
Result:
29
From which:
-(1.02198×10^-8)/((x(y^2)^(3/2))) = (3 sqrt(π))/4
Input interpretation:
Result:
Implicit plot:
30
Alternate form assuming x and y are real:
Alternate form:
Alternate form assuming x and y are positive:
Expanded forms:
Real solutions:
Solutions for the variable y:
31
Implicit derivatives:
From
We obtain:
Input:
32
Exact result:
Decimal approximation:
1.32934038…..
Property:
Series representations:
From
33
-(1.02198×10^-8)/((x(((0.000986822 - 0.00170923 i)/x^(1/3))^2)^(3/2)))
x = 0.5 :
Input interpretation:
Result:
Plots:
34
Alternate form:
Alternate form assuming x>0:
Alternate form assuming x is positive:
Expanded form:
Roots:
Series expansion at x = 0:
Series expansion at x = ∞:
35
Derivative:
Indefinite integral:
Definite integral after subtraction of diverging parts:
-(1.02198×10^-8)/((0.5(((0.000986822 - 0.00170923 i)/0.5^(1/3))^2)^(3/2)))
Input interpretation:
Result:
36
Polar coordinates:
1.32933
Polar forms:
From:
2
2
2
12
Γ
1+
4
2
=
2
2
32
2
Input:
Exact result:
Alternate form assuming m, ε, λ, and μ are real:
37
Alternate form:
Roots:
Root:
Series expansion at m = ∞:
Derivative:
38
Indefinite integral:
Limit:
Alternative representations:
39
From:
For:
λ = -5.18182 * 10^-9 μ = -1
0.00000000518182/2 *((-1)^(2*1/24)) (0.511^2 )^(1-2*1/24) Γ(-1+1/24)/(4π)^(2-
1/24)
Input interpretation:
Result:
Polar coordinates:
1.30355*10
-10
Polar forms:
40
Alternative representations:
Series representations:
41
Integral representations:
42
From
2
2
32
2
-5.18182 * 10^-9 * (-1)^(2*1/24) * (0.511^2)/(32Pi^2*1/24)
(((-0.00000000518182* (-1)^(2*1/24)))) * (0.511^2)/(32Pi^2*1/24)
Input interpretation:
Result:
43
Polar coordinates:
1.02822*10
-10
Polar forms:
Alternative representations:
Series representations:
44
Integral representations:
From which:
34/((((((ln((((-0.00000000518182* (-1)^(2*1/24)))) *
(0.511^2)/(32Pi^2*1/24))))))+e))
45
Input interpretation:
Result:
Polar coordinates:
1.6599 result that is very near to the 14th root of the following Ramanujan’s class
invariant =
505
/
101/5
3
= 1164.2696 i.e. 1.65578...
Indeed, from:
113+5
505
8
+
105+5
505
8
3
14
= 1,65578
46
Polar forms:
Alternative representations:
Series representations:
47
Integral representations:
48
From:
we obtain:
((((((-0.00000000518182* (-1)^(2*1/24)))) * 3/2*(0.511^2)))) / ((((1.02822×10^-10
)*2)))
Input interpretation:
Result:
Polar coordinates:
9.8696 = π
2
And:
1/6((((((-0.00000000518182* (-1)^(2*1/24)))) * 3/2*(0.511^2)))) / ((((1.02822×10^-
10 )*2)))
Input interpretation:
Result:
49
Polar coordinates:
1.64493 = ζ(2) =
2
6
Furthermore:
sqrt[((((((-0.00000000518182* (-1)^(2*1/24)))) * 3/2*(0.511^2)))) /
((((1.02822×10^-10 )*2)))]
Input interpretation:
Result:
Polar coordinates:
3.14159 = π
From:
(((-0.00000000518182* (-1)^(2*1/24)))) * (0.511^2)/(32Pi^2*1/24)
For ((((√(10-2√5) -2))⁄((√5-1)))) = 8πG; G = 0.011303146014
[2Pi((((√(10-2√5) -2))⁄((√5-1))))]/(0.011303146014)
50
we obtain:
0.932394 (((-0.00000000518182*(-1)^(2*1/24))))*(0.511^2)/(([4Pi((((√(10-2√5) -
2))⁄((√5-1)))^(2*1.65578))]/(0.011303146014))^2*1/24)
Input interpretation:
Result:
Polar coordinates:
1.02106*10
-10
Polar forms:
51
Series representations:
52
From which:
34/((ln+(0.932394 (((-0.00000000518182*(-
1)^(2*1/24))))*(0.511^2)/(([4Pi((((√(10-2√5) -2))⁄((√5-
1)))^(2*1.65578))]/(0.011303146014))^2*1/24) )+sqrt(2Pi)))
Input interpretation:
Result:
Polar coordinates:
1.64254 ≈ ζ(2) =
2
6
= 1.644934
53
Polar forms:
Alternative representations:
54
55
Series representations:
56
57
Integral representations:
59
For λ = -1
((-1/(1+(3/(16Pi^2))))) * ln[(((exp((16Pi^2)/(-3)))))]
Input:
60
Exact result:
Decimal approximation:
51.656533…
Property:
Alternate forms:
Alternative representations:
61
Series representations:
Integral representations:
62
We have also:
51.6565 1/ (((( integral_0^∞ 1/(1 + t^2) dt)^4) 1/(3 (3 + 64 ( integral_0^∞ 1/(1 + t^2)
dt)^2))))
Input interpretation:
Result:
4096
Computation result:
With regard 4096, we observe that:
From:
Modular equations and approximations to π – Srinivasa Ramanujan
Quarterly Journal of Mathematics, XLV, 1914, 350 – 372
63
We have that:
From:
we obtain:
e^(Pi*sqrt22)-24+(4096+276)*e^(-Pi*sqrt22) ≤ 64[(1+sqrt2)^12+(1-sqrt2)^12]
Input:
Result:
Difference:
(((e^(Pi*sqrt22)-24+(4096+276)*e^(-Pi*sqrt22)))) / [(1+sqrt2)^12+(1-sqrt2)^12]
Input:
64
Exact result:
Decimal approximation:
63.9999999….≈ 64
Property:
Alternate form:
Expanded forms:
65
Series representations:
66
67
68
and:
2*((((((e^(Pi*sqrt22)-24+(4096+276)*e^(-Pi*sqrt22)))) / [(1+sqrt2)^12+(1-
sqrt2)^12])))^2
Input:
Exact result:
Decimal approximation:
8191.99999…..≈ 8192
The total amplitude vanishes for gauge group SO(8192), while the vacuum energy is
negative and independent of the gauge group.
The vacuum energy and dilaton tadpole to lowest non-trivial order for the open
bosonic string. While the vacuum energy is non-zero and independent of the gauge
group, the dilaton tadpole is zero for a unique choice of gauge group, SO(2
13
) i.e.
SO(8192). (From: “Dilaton Tadpole for the Open Bosonic String “ Michael R.
Douglas and Benjamin Grinstein - September 2,1986)
Property:
69
Alternate form:
Expanded forms:
70
Series representations:
71
72
From the previous expression, we obtain:
27*sqrt(((51.6565 1/ (((( integral_0^∞ 1/(1 + t^2) dt)^4) 1/(3 (3 + 64 ( integral_0^∞
1/(1 + t^2) dt)^2)))))))+1
Input interpretation:
Result:
1729
73
This result is very near to the mass of candidate glueball f
0
(1710) scalar meson.
Furthermore, 1728 occurs in the algebraic formula for the j-invariant of an elliptic
curve. The number 1728 is one less than the Hardy–Ramanujan number 1729
(taxicab number)
Computation result:
[27*sqrt(((51.6565 1/ (((( integral_0^∞ 1/(1 + t^2) dt)^4) 1/(3 (3 + 64 ( integral_0^∞
1/(1 + t^2) dt)^2)))))))+1]^1/15
Input interpretation:
Result:
1.64382 ≈ ζ(2) =
2
6
= 1.644934
Computation result:
74
For ((((√(10-2√5) -2))⁄((√5-1)))) = 8πG; G = 0.011303146014
[2Pi((((√(10-2√5) -2))⁄((√5-1))))]/(0.011303146014)
We obtain:
-1/2(((((1/(1+(3/([0.9323942*Pi(1/(((√(10-2√5) -2))⁄((√5-
1))))]/(0.011303146014)))))) * ln[(((exp(([0.932394*2Pi(1/(((√(10-2√5) -2))⁄((√5-
1))))]/(0.011303146014))/(-3)))))])))
Input interpretation:
Result:
54.01104453….
75
Alternative representations:
76
77
From which:
-89/(((1/2(((((1/(1+(3/([0.9323942*Pi(1/(((√(10-2√5) -2))⁄((√5-
1))))]/(0.011303146014)))))) * ln[(((exp(([0.932394*2Pi(1/(((√(10-2√5) -2))⁄((√5-
1))))]/(0.011303146014))/(-3)))))]))))))
Input interpretation:
78
Result:
1.64781112383…. ≈ ζ(2) =
2
6
= 1.644934
And:
-89/(((1/2(((((1/(1+(3/([0.9323942*Pi(1/(((√(10-2√5) -2))⁄((√5-
1))))]/(0.011303146014)))))) * ln[(((exp(([0.932394*2Pi(1/(((√(10-2√5) -2))⁄((√5-
1))))]/(0.011303146014))/(-2-(0.937))))))]))))))
Where 0.937 is the following Omega meson Regge slope value:
Input interpretation:
Result:
1.61320709…. result that is a very good approximation to the value of the golden
ratio 1.618033988749...
79
Alternative representations:
80
81
Mathematical connections with some sectors of String Theory
From:
Modular equations and approximations to - Srinivasa Ramanujan
Quarterly Journal of Mathematics, XLV, 1914, 350 – 372
We have that:
82
From:
An Update on Brane Supersymmetry Breaking
J. Mourad and A. Sagnotti - arXiv:1711.11494v1 [hep-th] 30 Nov 2017
From the following vacuum equations:
we have obtained, from the results almost equals of the equations, putting
instead of
a new possible mathematical connection between the two exponentials. Thence, also
the values concerning p, C, β
E
and correspond to the exponents of e (i.e. of exp).
Thence we obtain for p = 5 and β
E
= 1/2:
6+
= 4096
18
Therefore, with respect to the exponentials of the vacuum equations, the Ramanujan’s
exponential has a coefficient of 4096 which is equal to 642, while -6C+ is equal to -
18. From this it follows that it is possible to establish mathematically, the dilaton
value.
For
83
exp((-Pi*sqrt(18)) we obtain:
Input:
Exact result:
Decimal approximation:
1.6272016… * 10
-6
Property:
Series representations:
Now, we have the following calculations:
6+
= 4096
18
18
= 1.6272016… * 10^-6
84
from which:
1
4096
6+
= 1.6272016… * 10^-6
0.000244140625
6+
=
18
= 1.6272016… * 10^-6
Now:
ln
18
= 13.328648814475 =
18
And:
(1.6272016* 10^-6) *1/ (0.000244140625)
Input interpretation:
Result:
0.006665017...
Thence:
0.000244140625
6+
=
18
Dividing both sides by 0.000244140625, we obtain:
85
0.000244140625
0.000244140625
6+
=
1
0.000244140625
18
6+
= 0.0066650177536
((((exp((-Pi*sqrt(18)))))))*1/0.000244140625
Input interpretation:
Result:
0.00666501785…
Series representations:
Now:
86
6+
= 0.0066650177536
=
= 0.00666501785…
From:
ln(0.00666501784619)
Input interpretation:
Result:
-5.010882647757…
Alternative representations:
Series representations:
87
Integral representation:
In conclusion:
6+ = 5.010882647757
and for C = 1, we obtain:
= 5.010882647757 + 6 = . =
Note that the values of n
s
(spectral index) 0.965, of the average of the Omega mesons
Regge slope 0.987428571 and of the dilaton 0.989117352243, are also connected to
the following two Rogers-Ramanujan continued fractions:
88
(http://www.bitman.name/math/article/102/109/)
The mean between the two results of the above Rogers-Ramanujan continued
fractions is 0.97798855285, value very near to the ψ Regge slope 0.979:
Also performing the 512
th
root of the inverse value of the Pion meson rest mass
139.57, we obtain:
((1/(139.57)))^1/512
Input interpretation:
Result:
0.99040073.... result very near to the dilaton value . = and to
the value of the following Rogers-Ramanujan continued fraction:
89
From
AdS Vacua from Dilaton Tadpoles and Form Fluxes - J. Mourad and A. Sagnotti
- arXiv:1612.08566v2 [hep-th] 22 Feb 2017 - March 27, 2018
We have:
For
ξ = 1
we obtain:
(2*e^(0.989117352243/2)) / (1+sqrt(((1-1/3*16/(Pi)^2*e^(2*0.989117352243)))))
Input interpretation:
90
Result:
Polar coordinates:
1.65919106525….. result very near to the 14th root of the following Ramanujan’s
class invariant =
505
/
101/5
3
= 1164.2696 i.e. 1.65578...
Series representations:
91
From
We obtain:
e^(4*0.989117352243) / (((1+sqrt(1-1/3*16/(Pi)^2*e^(2*0.989117352243)))))^7
[42(1+sqrt(1-
1/3*16/(Pi)^2*e^(2*0.989117352243)))+5*16/(Pi)^2*e^(2*0.989117352243)]
Input interpretation:
Result:
Polar coordinates:
54.76072411…..
92
Series representations:
93
From which:
e^(4*0.989117352243) / (((1+sqrt(1-1/3*16/(Pi)^2*e^(2*0.989117352243)))))^7
[42(1+sqrt(1-
1/3*16/(Pi)^2*e^(2*0.989117352243)))+5*16/(Pi)^2*e^(2*0.989117352243)]*1/34
Input interpretation:
Result:
Polar coordinates:
1.610609533…. result that is a good approximation to the value of the golden ratio
1.618033988749...
94
Series representations:
95
Now, we have:
For:
ξ = 1
= 0.989117352243
96
From
we obtain:
((2*e^(-0.989117352243/2))) /
((((1+sqrt(((1+1/3*(4Pi^2)/25*e^(2*0.989117352243))))))))
Input interpretation:
Result:
0.382082347529….
Series representations:
97
From which:
1+1/(((4((2*e^(-0.989117352243/2))) /
((((1+sqrt(((1+1/3*(4Pi^2)/25*e^(2*0.989117352243)))))))))))
Input interpretation:
Result:
1.6543092….. We note that, the result 1.6543092... is very near to the 14th root of the
following Ramanujan’s class invariant =
505
/
101/5
3
= 1164.2696 i.e.
1.65578...
Indeed:
98
113+5
505
8
+
105+5
505
8
3
14
= 1,65578
Series representations:
99
And from
we obtain:
e^(-4*0.989117352243) / [1+sqrt(((1+1/3*(4Pi^2)/25*e^(2*0.989117352243)))]^7 *
[42(1+sqrt(((1+1/3*(4Pi^2)/25*e^(2*0.989117352243)))-
13*(4Pi^2)/25*e^(2*0.989117352243)]
Input interpretation:
Result:
-0.034547055658…
100
Series representations:
101
From which:
47 *1/(((-1/(((((e^(-4*0.989117352243) /
[1+sqrt(((1+1/3*(4Pi^2)/25*e^(2*0.989117352243))))]^7 *
[42(1+sqrt(((1+1/3*(4Pi^2)/25*e^(2*0.989117352243))))-
13*(4Pi^2)/25*e^(2*0.989117352243))]))))))))
Input interpretation:
102
Result:
1.6237116159…. result that is an approximation to the value of the golden ratio
1.618033988749...
Series representations:
103
104
And again:
32((((e^(-4*0.989117352243) /
[1+sqrt(((1+1/3*(4Pi^2)/25*e^(2*0.989117352243))))]^7 *
[42(1+sqrt(((1+1/3*(4Pi^2)/25*e^(2*0.989117352243))))-
13*(4Pi^2)/25*e^(2*0.989117352243))]))))
Input interpretation:
Result:
-1.1055057810….
We note that the result -1.1055057810…. is very near to the value of Cosmological
Constant, less 10
-52
, thence 1.1056, with minus sign
105
Series representations:
106
107
And:
-[32((((e^(-4*0.989117352243) /
[1+sqrt(((1+1/3*(4Pi^2)/25*e^(2*0.989117352243))))]^7 *
[42(1+sqrt(((1+1/3*(4Pi^2)/25*e^(2*0.989117352243))))-
13*(4Pi^2)/25*e^(2*0.989117352243))]))))]^5
Input interpretation:
Result:
1.651220569…. result very near to the 14th root of the following Ramanujan’s class
invariant =
505
/
101/5
3
= 1164.2696 i.e. 1.65578...
108
Series representations:
109
110
We obtain also:
-[32((((e^(-4*0.989117352243) /
[1+sqrt(((1+1/3*(4Pi^2)/25*e^(2*0.989117352243))))]^7 *
[42(1+sqrt(((1+1/3*(4Pi^2)/25*e^(2*0.989117352243))))-
13*(4Pi^2)/25*e^(2*0.989117352243))]))))]^1/2
Input interpretation:
Result:
Polar coordinates:
1.05143035007
111
Series representations:
112
113
1 / -[32((((e^(-4*0.989117352243) /
[1+sqrt(((1+1/3*(4Pi^2)/25*e^(2*0.989117352243))))]^7 *
[42(1+sqrt(((1+1/3*(4Pi^2)/25*e^(2*0.989117352243))))-
13*(4Pi^2)/25*e^(2*0.989117352243))]))))]^1/2
Input interpretation:
Result:
Polar coordinates:
0.95108534763
We know that the primordial fluctuations are consistent with Gaussian purely
adiabatic scalar perturbations characterized by a power spectrum with a spectral
index n
s
= 0.965 ± 0.004, consistent with the predictions of slow-roll, single-field,
inflation.
Thence 0.95108534763 is a result very near to the spectral index n
s
, to the mesonic
Regge slope, to the inflaton value at the end of the inflation 0.9402 and to the value
of the following Rogers-Ramanujan continued fraction:
114
Series representations:
115
116
From the previous expression
= -0.034547055658…
we have also:
117
1+1/(((4((2*e^(-0.989117352243/2))) /
((((1+sqrt(((1+1/3*(4Pi^2)/25*e^(2*0.989117352243))))))))))) + (-0.034547055658)
Input interpretation:
Result:
1.61976215705….. result that is a very good approximation to the value of the golden
ratio 1.618033988749...
Series representations:
118
From
Properties of Nilpotent Supergravity
E. Dudas, S. Ferrara, A. Kehagias and A. Sagnotti - arXiv:1507.07842v2 [hep-th] 14
Sep 2015
We have that:
We analyzing the following equation:
119
We have:
(M^2)/3*[1-(b/euler number * k/sqrt6) * (φ- sqrt6/k) * exp(-(k/sqrt6)(φ- sqrt6/k))]^2
i.e.
V = (M^2)/3*[1-(b/euler number * k/sqrt6) * (φ- sqrt6/k) * exp(-(k/sqrt6)(φ-
sqrt6/k))]^2
For k = 2 and φ = 0.9991104684, that is the value of the scalar field that is equal to
the value of the following Rogers-Ramanujan continued fraction:
we obtain:
V = (M^2)/3*[1-(b/euler number * 2/sqrt6) * (0.9991104684- sqrt6/2) * exp(-
(2/sqrt6)(0.9991104684- sqrt6/2))]^2
120
Input interpretation:
Result:
Solutions:
Alternate forms:
Expanded form:
121
Alternate form assuming b, M, and V are positive:
Alternate form assuming b, M, and V are real:
Derivative:
Implicit derivatives:
122
Global minimum:
Global minima:
123
From:
we obtain
(225.913 (-0.054323 M^2 + 6.58545×10^-10 sqrt(M^4)))/M^2
Input interpretation:
Result:
Plots:
124
Alternate form assuming M is real:
-12.2723 result very near to the black hole entropy value 12.1904 = ln(196884)
Alternate forms:
Expanded form:
Property as a function:
Parity
Series expansion at M = 0:
125
Series expansion at M = ∞:
Derivative:
Indefinite integral:
Global maximum:
Global minimum:
126
Limit:
Definite integral after subtraction of diverging parts:
From b that is equal to
from:
Result:
we obtain:
1/3 (0.0814845 ((225.913 (-0.054323 M^2 + 6.58545×10^-10 sqrt(M^4)))/M^2 ) +
1)^2 M^2
127
Input interpretation:
Result:
Plots: (possible mathematical connection with an open string)
M = -0.5; M = 0.2
(possible mathematical connection with an open string)
M = 2 ; M = 3
128
Root:
Property as a function:
Parity
Series expansion at M = 0:
Series expansion at M = ∞:
Definite integral after subtraction of diverging parts:
For M = - 0.5 , we obtain:
129
1/3 (0.0814845 ((225.913 (-0.054323 (-0.5)^2 + 6.58545×10^-10 sqrt((-0.5)^4)))/(-
0.5)^2 ) + 1)^2 * (-0.5^2)
Input interpretation:
Result:
-4.38851344947*10
-16
For M = 0.2:
1/3 (0.0814845 ((225.913 (-0.054323 0.2^2 + 6.58545×10^-10 sqrt(0.2^4)))/0.2^2 ) +
1)^2 0.2^2
130
Input interpretation:
Result:
7.021621519159*10
-17
For M = 3:
1/3 (0.0814845 ((225.913 (-0.054323 3^2 + 6.58545×10^-10 sqrt(3^4)))/3^2 ) + 1)^2
3^2
Input interpretation:
Result:
1.57986484181*10
-14
131
For M = 2:
1/3 (0.0814845 ((225.913 (-0.054323 2^2 + 6.58545×10^-10 sqrt(2^4)))/2^2 ) + 1)^2
2^2
Input interpretation:
Result:
7.021621519*10
-15
From the four results
7.021621519*10^-15 ; 1.57986484181*10^-14 ; 7.021621519159*10^-17 ;
-4.38851344947*10^-16
we obtain, after some calculations:
132
sqrt[1/(2Pi)(7.021621519*10^-15 + 1.57986484181*10^-14 +7.021621519*10^-17 -
4.38851344947*10^-16)]
Input interpretation:
Result:
5.9776991059*10
-8
result very near to the Planck's electric flow 5.975498 × 10
−8
that
is equal to the following formula:
We note that:
1/55*(([(((1/[(7.021621519*10^-15 + 1.57986484181*10^-14 +7.021621519*10^-17
-4.38851344947*10^-16)])))^1/7]-((log^(5/8)(2))/(2 2^(1/8) 3^(1/4) e log^(3/2)(3)))))
Input interpretation:
133
Result:
1.6181818182… result that is a very good approximation to the value of the golden
ratio 1.618033988749...
From the Planck units:
Planck Length
5.729475 * 10
-35
Lorentz-Heaviside value
Planck’s Electric field strength
1.820306 * 10
61
V*m Lorentz-Heaviside value
Planck’s Electric flux
5.975498*10
-8
V*m Lorentz-Heaviside value
134
Planck’s Electric potential
1.042940*10
27
V Lorentz-Heaviside value
Relationship between Planck’s Electric Flux and Planck’s Electric Potential
E
P
* l
P
= (1.820306 * 10
61
) * 5.729475 * 10
-35
Input interpretation:
Result:
Scientific notation:
1.042939771935*10
27
≈ 1.042940*10
27
Or:
E
P
* l
P
2
/ l
P
= (5.975498*10
-8
)*1/(5.729475 * 10
-35
)
Input interpretation:
Result:
1.042939885417*10
27
≈ 1.042940*10
27
135
Observations
We note that, from the number 8, we obtain as follows:
We notice how from the numbers 8 and 2 we get 64, 1024, 4096 and 8192, and that 8
is the fundamental number. In fact 8
2
= 64, 8
3
= 512, 8
4
= 4096. We define it
"fundamental number", since 8 is a Fibonacci number, which by rule, divided by the
previous one, which is 5, gives 1.6 , a value that tends to the golden ratio, as for all
numbers in the Fibonacci sequence
136
“Golden” Range
Finally we note how 8
2
= 64, multiplied by 27, to which we add 1, is equal to 1729,
the so-called "Hardy-Ramanujan number". Then taking the 15th root of 1729, we
obtain a value close to ζ(2) that 1.6438 ..., which, in turn, is included in the range of
what we call "golden numbers"
Furthermore for all the results very near to 1728 or 1729, adding 64 = 8
2
, one obtain
values about equal to 1792 or 1793. These are values almost equal to the Planck
multipole spectrum frequency 1792.35 and to the hypothetical Gluino mass
Acknowledgments
We would like to thank Professor Augusto Sagnotti theoretical physicist at Scuola
Normale Superiore (Pisa – Italy) for his very useful explanations and his availability
137
References
Complex Analysis in Number Theory – 22.11.1994 - Anatoly A. Karatsuba
On the Zeros of the Davenport Heilbronn Function
S. A. Gritsenko - Received May 15, 2016 - ISSN 0081-5438, Proceedings of the
Steklov Institute of Mathematics, 2017, Vol. 296, pp. 65–87.
Course of Field Theory and Gravity - Prof. Augusto Sagnotti (SNS Pisa-Italy)
Field Theory: A Modern Primer - By Pierre Ramond - Copyright Year 1997
Modular equations and approximations to - Srinivasa Ramanujan
Quarterly Journal of Mathematics, XLV, 1914, 350 – 372
An Update on Brane Supersymmetry Breaking
J. Mourad and A. Sagnotti - arXiv:1711.11494v1 [hep-th] 30 Nov 2017
AdS Vacua from Dilaton Tadpoles and Form Fluxes - J. Mourad and A. Sagnotti
- arXiv:1612.08566v2 [hep-th] 22 Feb 2017 - March 27, 2018
Properties of Nilpotent Supergravity
E. Dudas, S. Ferrara, A. Kehagias and A. Sagnotti - arXiv:1507.07842v2 [hep-th] 14
Sep 2015