1
On the study of various equations concerning the Plateau Problem and the
Inflation. Possible mathematical connections with some sectors of Number
Theory, String Theory and some cosmological parameters.
Michele Nardelli
1
, Antonio Nardelli
2
Abstract
In this paper, we analyze various equations concerning the Plateau Problem and the
Inflation. We describe the new possible mathematical connections with some sectors
of Number Theory, String Theory and cosmological parameters
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
3
Introduction
In 1983, it was shown that inflation could be eternal, leading to a multiverse in which
space is broken up into bubbles or patches whose properties differ from patch to
patch spanning all physical possibilities.
When the false vacuum decays, the lower-energy true vacuum forms through a
process known as bubble nucleation. In this process, instanton effects cause a
bubble containing the true vacuum to appear. The walls of the bubble (or domain
walls) have a positive surface tension, as energy is expended as the fields roll over
the potential barrier to the true vacuum.
In mathematics, a ball is the space bounded by a sphere. It may be a closed ball
(including the boundary points that constitute the sphere) or an open ball (excluding
them). (From Wikipedia)
We propose that some equations concerning the “balls”, thus various sectors and
theorems of Measure Theory, can be related with several parameters of some
cosmological models as the “Multiverse” and the “Eternal Inflation” linked to it,
which provides that space is divided into bubbles or patches whose properties differ
from patch to patch and spanning all physical possibilities.
5
Geometric figure
Alternate forms
Expanded form
Real root
6
Complex roots
Polynomial discriminant
Properties as a real function
Domain
Range
Derivative
Indefinite integral
Global minimum
7
Definite integral after subtraction of diverging parts
(((1*10^-5)^2*(0.206897 + 1.84539 i)^2))/2 (1-0.12*(0.206897 + 1.84539
i)+0.29*(0.206897 + 1.84539 i)^2)
Input interpretation
Result
Polar coordinates
V = 2.83672*10
-16
(((1*10^-5)^2*(0.206897 - 1.84539 i)^2))/2 (1-0.12*(0.206897 - 1.84539
i)+0.29*(0.206897 - 1.84539 i)^2)
Input interpretation
Result
8
Polar coordinates
2.83672*10
-16
Or, considering
(0.206897 + 1.84539 i)
Input interpretation
Result
Polar coordinates
1.85695
(((1*10^-5)^2*(1.85695)^2))/2 (1-0.12*(1.85695)+0.29*(1.85695)^2)
Input interpretation
Result
3.06406187…*10
-10
For:
9
sqrt6*cosh(x/(sqrt6))
Input
Plots (figures that can be related to the open strings)
10
Alternate form
Roots
Properties as a real function
Domain
Range
Parity
Periodicity
11
Series expansion at x=0
Derivative
Indefinite integral
Global minimum
Alternative representations
12
Series representations
Integral representations
Definite integral mean square
13
sqrt6*sinh(x/(sqrt6))
Input
Plots (figures that can be related to the open strings)
Alternate form
Roots
Integer root
14
Properties as a real function
Domain
Range
Bijectivity
Parity
Periodicity
Series expansion at x=0
Derivative
15
Indefinite integral
Alternative representations
Series representations
16
Integral representations
Definite integral over a half-period
Definite integral mean square
From:
2.83672*10^-16 * tanh^2(x/(sqrt6)y)
Input interpretation
17
Result
3D plot (figure that can be related to a D-brane/Instanton)
Contour plot
Alternate forms
18
Real root
Properties as a function
Domain
Range
Root for the variable y
19
Series expansion at x=0
Partial derivatives
Indefinite integral
Global minimum
Limit
20
2.83672×10^-16 tanh^2(2/(sqrt(6) sqrt(-2)))
Input interpretation
Result
-1.20364…*10
-16
From:
For: φ = 2 ; α = - 2
2.83672×10^-16 (1-exp(sqrt(2/(-6))*2))^2
Input interpretation
Result
21
Polar coordinates
3.38028*10
-16
From:
2.83672×10^-16 (1- (16/(1*10^-5)^4))
Input interpretation
Result
-453875.19999….
From:
Sqrt(-12) tanh(2/(sqrt(-12))
Input
22
Exact result
Decimal approximation
= 2.25647383035……
Property
Alternate forms
Alternative representations
23
Series representations
Integral representations
24
From:
1/2*sqrt(-1) [(∂_u*sqrt6*∂^u*sqrt6-
∂_u*2.25647383035*∂^u*2.25647383035)+1/6*(6-2.25647383035^2)*x-
1/18(2.25647383035^2-6)^2*y(2.25647383035/(sqrt6))]
1/2*sqrt(-1) [(sqrt6*sqrt6-2.25647383035*2.25647383035)+1/6*(6-
2.25647383035^2)*x-1/18(2.25647383035^2-6)^2*y(2.25647383035/(sqrt6))]
Input interpretation
Result
3D plots
Real part (figures that can be related to the D-branes/Instantons)
25
Imaginary part
26
Contour plots
Real part
27
Imaginary part
28
Alternate forms
Expanded form
Alternate form assuming x and y are real
Real root
Root
Properties as a function
Domain
Range
Partial derivatives
29
Indefinite integral
Definite integral over a disk of radius R
Definite integral over a square of edge length 2 L
From:
For x = R(g) = 5 and y = F = 20 :
1/2 i (0.15138764216*5 - 0.042224597800*20 + 0.90832585295)
Input interpretation
30
Result
Polar coordinates
0.4103860539
From:
Sqrt(-1) [1/2*(sqrt6 * sqrt6)+6/12*5-x/4*(sqrt6)^4]
Input
Result
31
Plot
Alternate forms
Expanded form
Root
Properties as a real function
Domain
Range
32
Derivative
Indefinite integral
Thence, for x = 0.9 :
Sqrt(-1) [1/2*(sqrt6 * sqrt6)+6/12*5-(0.9)/4*(sqrt6)^4]
Input
Result
Polar coordinates
2.6
From:
33
Sqrt(-1)*(5/2-9*0.9)
Input
Result
Polar coordinates
5.6
From the above two expressions, after some calculations, we obtain:
64((Sqrt(-1)*(5/2-9*0.9))-(Sqrt(-1) [1/2*(sqrt6 * sqrt6)+6/12*5-
(0.9)/4*(sqrt6)^4]))^3+1i
Input
Result
Polar coordinates
1729
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. (1728 = 8
2
* 3
3
) The number 1728 is one less than the Hardy–Ramanujan
number 1729 (taxicab number)
34
(64((Sqrt(-1)*(5/2-9*0.9))-(Sqrt(-1) [1/2*(sqrt6 * sqrt6)+6/12*5-
(0.9)/4*(sqrt6)^4]))^3+1i)^1/15
Input
Result
Polar coordinates
1.64382 ≈ ζ(2) =
2
6
= 1.644934 (trace of the instanton shape)
From:
Sqrt(-1) [1/2*5-1/2*2*2-20(tanh(2/(sqrt6))]
Input
Exact result
35
Decimal approximation
-12.963170378…i
Property
Polar coordinates
Polar forms
Approximate form
Alternate forms
36
Expanded form
Alternative representations
37
Series representations
Integral representation
From
we obtain:
-1/(6+1/e)(-1/4(i (1/2 - 20 tanh(sqrt(2/3)))))^2
Input
38
Exact result
Decimal approximation
1.64933032…. ≈ ζ(2) =
2
6
= 1.644934 (trace of the instanton shape)
Alternate forms
39
Expanded form
Alternative representations
Series representations
40
Integral representation
From:
1/2*sqrt(-1) [(sqrt6 * sqrt6 - 2.25647383035*2.25647383035)+1/6*(6-
2.25647383035^2)*5-0.9*1/4*(2.25647383035^2-6)^2]
Input interpretation
Result
41
Polar coordinates
0.739813
From the algebraic sum between the two previous expressions
and:
we obtain, after some calculations:
2((1/2*sqrt(-1) [(sqrt6 * sqrt6 - 2.25647383035*2.25647383035)+1/6*(6-
2.25647383035^2)*5-0.9*1/4*(2.25647383035^2-6)^2]-1/2 i (0.15138764216*5 -
0.042224597800*20 + 0.90832585295)))
Input interpretation
Result
42
Polar coordinates
0.658854
From which:
1+0.658854
Input interpretation
Result
1.658854 result very near to the 14th root of the following Ramanujan’s class
invariant =
505
/
101/5
3
= 1164.2696 i.e. 1.65578...
From:
1/2 i (0.15138764216*5 - 0.042224597800*20 + 0.90832585295)
and:
43
sqrt((((((2.83672×10^-16 (1-exp(sqrt(2/(-6))*2))^2)) / ((2.83672×10^-16
tanh^2(2/(sqrt(6) sqrt(-2)))))))-((1/4 i (0.15138764216*5 - 0.042224597800*20 +
0.90832585295)))))
Input interpretation
Result
Polar coordinates
1.61926 result that is a very good approximation to the value of the golden ratio
1.618033988749...
From:
For R(g) = 5 ; = 2.25647383035 ; F = 20
44
Sqrt(-1) [1/2*5(1-1/6*2.25647383035^2)-1/2*2.25647383035*2.25647383035-
20(2.25647383035/(sqrt6))(1/6*2.25647383035^2-1)^2]
Input interpretation
Result
Polar coordinates
2.5896139461
From which:
sqrt((Sqrt(-1) [1/2*5(1-1/6*2.25647383035^2)-1/2*2.25647383035*2.25647383035-
20(2.25647383035/(sqrt6))(1/6*2.25647383035^2-1)^2]))
Input interpretation
Result
45
Polar coordinates
1.6092277484 result quite near to the value of the golden ratio 1.618033988749...
From:
1/(1-(2.25647383035^2)/6)
Input interpretation
Result
6.6055589858448…
From:
20(tanh(2/sqrt(6)))
Input
46
Exact result
Decimal approximation
13.463170378839……
Property
Alternate forms
47
Alternative representations
Series representations
48
Integral representation
From the above two expressions, after some calculations, we obtain:
34*1/((1/(1-(2.25647383035^2)/6))+(20(tanh(2/sqrt(6))))+1)
Input interpretation
Result
1.6137660422…. result that is a very good approximation to the value of the golden
ratio 1.618033988749...
Alternative representations
49
Series representations
Integral representation
50
From:
For φ = 2 ; α = - 2 ; R(g) = 5 ; = 2.25647383035 ; m = 10
-5
1/2*5-1/2*(2.25647383035^2)/(1+(2.25647383035^2)/(12))^2-1/2*(1*10^-
5)^2*2.25647383035^2
Input interpretation
Result
1.2450573192338…..
From which:
1+1/((1/2*5-1/2*(2.25647383035^2)/(1+(2.25647383035^2)/(12))^2-1/2*(1*10^-
5)^2*2.25647383035^2))^2
Input interpretation
51
Result
1.64509148….. ≈ ζ(2) =
2
6
= 1.644934 (trace of the instanton shape)
For: φ = 2 ; α = - 2 ; m = 10^-5 ; = 2.25647383035 ;
From:
-6*(1*10^-5)^2*tanh^2(2/(sqrt(-12)))
Input interpretation
Result
Decimal approximation
2.545837073…*10
-10
52
Property
Alternate forms
From:
For: φ = 2 ; α = - 2 ; m = 10^-5 ; = 2.25647383035 ;
1
2
2
1
2
6
2
.
We consider = x:
53
1/2 (partial derivative(x^2))/(1-x^2/(6α))^2
Input interpretation
Result
3D plot (figure that can be related to a D-brane/Instanton)
Contour plot
54
Alternate form
Partial fraction expansion
Expanded form
Roots
Properties as a real function
Domain
Range
55
Parity
Series expansion at x=0
Series expansion at x=∞
Derivative
Indefinite integral
Limit
56
From:
-(108 α^2 (x^2 + 2 α))/(x^2 - 6 α)^3
Input
3D plot (figure that can be related to a D-brane/Instanton)
Contour plot
57
Alternate forms
Partial fraction expansion
Expanded forms
Roots
58
Integer roots
Properties as a real function
Domain
Range
Parity
Roots for the variable x
59
Series expansion at x=0
Series expansion at x=∞
Derivative
Indefinite integral
Limit
Series representations
60
From:
((-108 x^2 - 216 α) α^2)/(x^6 + α (-18 x^4 + (108 x^2 - 216 α) α))
Input
Result
3D plot (figure that can be related to a D-brane/Instanton)
61
Contour plot
Alternate forms
Partial fraction expansion
Expanded forms
62
Roots
Integer roots
Properties as a real function
Domain
Range
63
Parity
Roots for the variable x
Series expansion at x=0
Series expansion at x=∞
Derivative
Indefinite integral
64
Limit
Series representations
From:
For α = -2 :
65
((-108 x^2 - 216(-2)) (-2)^2)/(x^6 + (-2)(-18 x^4 + (108 x^2 - 216(-2)) (-2)))
Input
Result
Plots (figures that can be related to the open strings)
Alternate forms
66
Partial fraction expansion
Expanded forms
Roots
67
Properties as a real function
Domain
Range
Parity
Series expansion at x=0
Series expansion at x=∞
Derivative
Indefinite integral
68
Global maximum
Global minima
Limit
Definite integral
Series representations
69
From:
For x = 5.1 :
1728/(5.1^6 - 2 (-18 5.1^4 - 2 (432 + 108 5.1^2))) - (432 5.1^2)/(5.1^6 - 2 (-18 5.1^4
- 2 (432 + 108 5.1^2)))
Input
Result
-0.173145090039….
For x = 25.3 :
1728/(25.3^6 - 2 (-18 25.3^4 - 2 (432 + 108 25.3^2))) - (432 25.3^2)/(25.3^6 - 2 (-18
25.3^4 - 2 (432 + 108 25.3^2)))
70
Input
Result
-0.0009910137819….
From the two previous expressions, after some calculations, we obtain:
-1/2((-0.0009910137819429+((1728/(5.1^6 - 2 (-18 5.1^4 - 2 (432 + 108 5.1^2))) -
(432 5.1^2)/(5.1^6 - 2 (-18 5.1^4 - 2 (432 + 108 5.1^2)))))))
Input interpretation
Result
0.08706805191….
From which:
19*((-1/2((-0.0009910137819429+((1728/(5.1^6 - 2 (-18 5.1^4 - 2 (432 + 108
5.1^2))) - (432 5.1^2)/(5.1^6 - 2 (-18 5.1^4 - 2 (432 + 108 5.1^2)))))))))
where 19 is a Twin prime number
71
Input interpretation
Result
1.6542929863…. ≈ ζ(2) =
2
6
= 1.644934 (trace of the instanton shape)
From:
For N = 60:
1-2/60
Input
Exact result
72
Decimal approximation
0.966666…. = n
s
And:
-2*12/60^2
Input
Exact result
Decimal approximation
-0.0066666…. = r
Repeating decimal
Instead of N = 60 , if we consider N = 64, we obtain:
1-2/64
Input
73
Exact result
Decimal form
0.96875 = n
s
We note that, from:
1-(e^-Pi/(1+e^(-2Pi)/(1+e^(-3Pi)/(1+e^(-4Pi)))))
Input
Decimal approximation
0.956866624345….
Property
74
Alternate forms
Alternative representations
75
Series representations
Integral representations
76
And from:
1-e^(-Pi*sqrt5)/(1+e^(-2Pi*sqrt5)/(1+e^(-3Pi*sqrt5)/(1+e^(-4Pi*sqrt5))))
Input
Exact result
77
Decimal approximation
0.999110468396….
Alternate forms
78
Series representations
79
performing the algebraic sum between the two above Rogers-Ramanujan continued
fractions, we obtain, after some calculations:
80
sqrt((3 log(π))/14)*[((1-e^(-Pi*sqrt5)/(1+e^(-2Pi*sqrt5)/(1+e^(-3Pi*sqrt5)/(1+e^(-
4Pi*sqrt5)))))) + ((1-(e^-Pi/(1+e^(-2Pi)/(1+e^(-3Pi)/(1+e^(-4Pi)))))))]
where
Input
Exact result
Decimal approximation
0.9687503767….that is the value of
= 1
2
81
Alternate forms
82
Expanded form
Alternative representations
83
From:
we perform the following double integration:
((-(e^(-π) √((3 log(π))/14))/(1 + e^(-2 π)/(1 + e^(-3 π)/(1 + e^(-4 π)))) - (e^(-√(5) π)
√((3 log(π))/14))/(1 + e^(-2 √(5) π)/(1 + e^(-3 √(5) π)/(1 + e^(-4 √(5) π)))) + √((6
log(π))/7)))dxdy
and we obtain:
Indefinite integral
84
3D plot (figure that can be related to a D-brane/Instanton)
Contour plot
85
Definite integral over a disk of radius R
Definite integral over a square of edge length 2 L
86
From the result of
we obtain:
π R^2 (-(e^(-π) √((3 log(π))/14))/(1 + e^(-2 π)/(1 + e^(-3 π)/(1 + e^(-4 π)))) - (e^(-
√(5) π) √((3 log(π))/14))/(1 + e^(-2 √(5) π)/(1 + e^(-3 √(5) π)/(1 + e^(-4 √(5) π)))) +
√((6 log(π))/7))
Input
Exact result
87
Plot (figure that can be related to an open string)
Geometric figure
88
Alternate forms
89
Expanded form
Polynomial discriminant
Derivative
If we consider R = 1 :
π (-(e^(-π) √((3 log(π))/14))/(1 + e^(-2 π)/(1 + e^(-3 π)/(1 + e^(-4 π)))) - (e^(-sqrt(5)
π) √((3 log(π))/14))/(1 + e^(-2 √(5) π)/(1 + e^(-3 √(5) π)/(1 + e^(-4 √(5) π)))) + √((6
log(π))/7))
Input
90
Exact result
Decimal approximation
3.0434190668….
Alternate forms
91
Expanded form
Alternative representations
92
From which, after some calculations:
1+1/(1/6((π (-(e^(-π) √((3 log(π))/14))/(1 + e^(-2 π)/(1 + e^(-3 π)/(1 + e^(-4 π)))) -
(e^(-√(5) π) √((3 log(π))/14))/(1 + e^(-2 √(5) π)/(1+e^(-3 √(5) π)/(1+e^(-4 √(5)
π))))+√((6 log(π))/7))))^2)
Input
93
Exact result
Decimal approximation
1.647780299…. ≈ ζ(2) =
2
6
= 1.644934 (trace of the instanton shape)
Alternate forms
94
Alternative representations
95
And again:
0.96875+1/(1/6((π (-(e^(-π) √((3 log(π))/14))/(1+e^(-2 π)/(1 + e^(-3 π)/(1+e^(-4
π)))) - (e^(-√(5) π) √((3 log(π))/14))/(1 + e^(-2 √(5) π)/(1+e^(-3 √(5) π)/(1+e^(-4
√(5) π))))+√((6 log(π))/7))))^2)
where 0.96875 = n
s
Input
Result
1.6165302997…. result that is a very good approximation to the value of the golden
ratio 1.618033988749...
96
Alternative representations
97
Series representations
98
99
Integral representations
100
From:
SHARP STABILITY INEQUALITIES FOR THE PLATEAU PROBLEM - G.
De Philippis & F. Maggi - j. differential geometry 96 (2014) 399-456
We have:
101
We consider: k = 2 ; h = 12 ; α = 1 ; β = 11 ; t = 0.5
(11^4*0.5^3-3*11^2*0.5^2-3*11*0.5+1) ≥ 1/64
Input
Result
Difference
1723.86
For k = 2 ; h = 12 ; α = 1 ; β = 11 ; t = 0.5
0.5^3-3*11*0.5^2-3*11^2*0.5+11^4 ≥ (11^4)/64
Input
Result
102
Difference
14222.6
From the left-hand side of the two inequality, we obtain:
(11^4*0.5^3-3*11^2*0.5^2-3*11*0.5+1)
Input
Result
1723.875
(0.5^3-3*11*0.5^2-3*11^2*0.5+11^4)
Input
Result
14451.375
From the ratio of the two expressions, after some calculations, we obtain:
1/(5+(1/π^(3/2)))[(0.5^3-3*11*0.5^2-3*11^2*0.5+11^4) / (11^4*0.5^3-
3*11^2*0.5^2-3*11*0.5+1)]
where
Input
103
Result
1.618483427…. result that is a very good approximation to the value of the golden
ratio 1.618033988749...
Alternative representations
Series representations
104
Integral representations
Multiplying the two expressions, after some calculations and raising to the 16
th
power, we get :
105
[(0.5^3-3*11*0.5^2-3*11^2*0.5+11^4)(11^4*0.5^3-3*11^2*0.5^2-3*11*0.5+1)]^16
* ((71/11 sqrt(71/11) π^4))
where
(71 and 11 are twin prime numbers)
Input
Result
0.35159…*10
122
≈ Λ
Q
The observed value of ρ
Λ
or Λ today is precisely the classical dual of its quantum
precursor values ρ
Q
, Λ
Q
in the quantum very early precursor vacuum U
Q
as
determined by our dual equations. With regard the Cosmological constant,
fundamental are the following results: Λ = 2.846 * 10
-122
and Λ
Q
= 0.3516 * 10
122
(New Quantum Structure of the Space-Time - Norma G. SANCHEZ - arXiv:1910.13382v1
[physics.gen-ph] 28 Oct 2019)
Series representations
106
Now, we have that:
[((3(x^(13/2) – y^(13/2))))-(((11/2*(x^(3/2)*y^(3/2)))))*(((x^(7/2)-y^(7/2))))] /
((((z^(15/2)))))
Input
Result
107
Alternate forms
Alternate form assuming x, y, and z are positive
Expanded form
Property as a function
Domain
Series expansion at x=0
108
Partial derivatives
Indefinite integral
From the result of the integral
we obtain:
109
(24 x^(15/2) - 55 x^6 y^(3/2) + 132 x^(5/2) y^5 - 180 x y^(13/2))/(60 z^(15/2))
Input
Alternate form
Expanded form
Alternate form assuming x, y, and z are positive
Property as a function
Domain
Partial derivatives
110
Indefinite integral
From:
for x = 4, y = 8 and z = 16 :
Input
Result
111
Decimal approximation
-0.0062048867….
Alternate form
Minimal polynomial
Expanded form
Inverting, after some calculations, we obtain:
(-1/2*(1/((((-55 4^6 8^(3/2) + 132 4^(5/2) 8^5 - 180 4 8^(13/2) + 24 4^7 sqrt(4/16)
sqrt(16))/(60 16^(15/2)))))+34-0.9568666373))^2-7-Φ
Input interpretation
Result
4096.07599456…. ≈ 4096 = 64
2
112
And, consequently:
27((-1/2*(1/((((-55 4^6 8^(3/2) + 132 4^(5/2) 8^5 - 180 4 8^(13/2) + 24 4^7
sqrt(4/16) sqrt(16))/(60 16^(15/2)))))+34-0.9568666373)))-Φ
Input interpretation
Result
1729.00416372…
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. (1728 = 8
2
* 3
3
) The number 1728 is one less than the Hardy–Ramanujan
number 1729 (taxicab number)
113
((27((-1/2*(1/((((-55 4^6 8^(3/2) + 132 4^(5/2) 8^5 - 180 4 8^(13/2) + 24 4^7
sqrt(4/16) sqrt(16))/(60 16^(15/2)))))+34-0.9568666373)))-Φ))^1/15
Input interpretation
Result
1.643815492654…. ≈ ζ(2) =
2
6
= 1.644934 (trace of the instanton shape)
From:
((16^6)/512)
Input
Result
32768
114
From:
c/(z^2) ((x/(sqrt(3))-y/(sqrt(3))))
Input
Result
Alternate forms
Expanded forms
115
Roots
Integer roots
Properties as a function
Domain
Range
Parity
116
Derivative
Indefinite integral
Limit
Series representations
117
From:
For
(sqrt3/16)*1/((8^3)^2) ((2^3/sqrt(3) - 4^3/sqrt(3)))
For x = 2
3
, y = 4
3
, z = 8
3
Input
Exact result
Decimal form
-0.0000133514404296875
118
From:
((11/2(((8^1.5 – 64^1.5)(8^5+64^5)-5/2(8^6.5-64^6.5)))))/((8^5+64^5)^1.5)
Input
Result
0.132695315997….
Dividing the two above expressions, we obtain:
(((((11/2(((8^1.5 – 64^1.5)(8^5+64^5)-5/2(8^6.5-64^6.5)))))/((8^5+64^5)^1.5)))) /
(((sqrt3/16)*1/((8^3)^2) ((2^3/sqrt(3) - 4^3/sqrt(3)))))
Input
Result
-9938.65169049….
119
From which:
[-(1/6*(((((11/2(((8^1.5 – 64^1.5)(8^5+64^5)-5/2(8^6.5-
64^6.5)))))/((8^5+64^5)^1.5)))) / (((sqrt3/16)*1/((8^3)^2) ((2^3/sqrt(3) -
4^3/sqrt(3)))))-72-Φ)]
Input
Result
1729.06….
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. (1728 = 8
2
* 3
3
) The number 1728 is one less than the Hardy–Ramanujan
number 1729 (taxicab number)
(1/27[-(1/6*(((((11/2(((8^1.5 – 64^1.5)(8^5+64^5)-5/2(8^6.5-
64^6.5)))))/((8^5+64^5)^1.5)))) / (((sqrt3/16)*1/((8^3)^2) ((2^3/sqrt(3) -
4^3/sqrt(3)))))-72)])^2-2
Input
120
Result
4096.10…. ≈ 4096 = 64
2
(([-(1/6*(((((11/2(((8^1.5 – 64^1.5)(8^5+64^5)-5/2(8^6.5-
64^6.5)))))/((8^5+64^5)^1.5)))) / (((sqrt3/16)*1/((8^3)^2) ((2^3/sqrt(3) -
4^3/sqrt(3)))))-72-Φ)]))^1/15
Input
Result
1.64382…. ≈ ζ(2) =
2
6
= 1.644934 (trace of the instanton shape)
From:
[3(8^6.5-64^6.5)-11/2(((((8)^1.5*(64)^1.5))*(8^3.5-64^3.5)))] / (((8^3)^7.5))
Input
121
Result
-7.26268…*10
-9
Dividing the two above expressions:
and
after some calculations, we obtain:
e^(-2 + 2/e - e) cot^2(e π) (((-((11/2(((8^1.5 – 64^1.5)(8^5+64^5)-5/2(8^6.5-
64^6.5)))))/((8^5+64^5)^1.5) *1/((([3(8^6.5-64^6.5)-
11/2(((((8)^1.5*(64)^1.5))*(8^3.5-64^3.5)))] / (((8^3)^7.5))))))))^17
Input
Result
0.3516…*10
122
≈ Λ
Q
122
The observed value of ρ
Λ
or Λ today is precisely the classical dual of its quantum
precursor values ρ
Q
, Λ
Q
in the quantum very early precursor vacuum U
Q
as
determined by our dual equations. With regard the Cosmological constant,
fundamental are the following results: Λ = 2.846 * 10
-122
and Λ
Q
= 0.3516 * 10
122
(New Quantum Structure of the Space-Time - Norma G. SANCHEZ - arXiv:1910.13382v1
[physics.gen-ph] 28 Oct 2019)
Alternative representations
123
Series representations
124
Integral representation
And again:
(((((-((11/2(((8^1.5 – 64^1.5)(8^5+64^5)-5/2(8^6.5-64^6.5)))))/((8^5+64^5)^1.5)
*1/((([3(8^6.5-64^6.5)-11/2(((((8)^1.5*(64)^1.5))*(8^3.5-64^3.5)))] /
(((8^3)^7.5))))))))))^1/35
Input
Result
1.612422379593…. result that is a very good approximation to the value of the
golden ratio 1.618033988749...
125
(((((-((11/2(((8^1.5 – 64^1.5)(8^5+64^5)-5/2(8^6.5-64^6.5)))))/((8^5+64^5)^1.5)
*1/((([3(8^6.5-64^6.5)-11/2(((((8)^1.5*(64)^1.5))*(8^3.5-64^3.5)))] /
(((8^3)^7.5))))))))))^1/33
Input
Result
1.6597905632…. result very near to the 14th root of the following Ramanujan’s class
invariant =
505
/
101/5
3
= 1164.2696 i.e. 1.65578...
From:
For: A = 2 ; B = 4 ; t = 0.5
Input
126
Result
2^4*0.5^3-3*2^2*4*0.5^2-3*2*4^2*0.5+4^4 ≥ (4^4)/64
Input
Result
(2^4*0.5^3-3*2^2*4*0.5^2-3*2*4^2*0.5+4^4)
Input
Result
198
Thence:
From:
127
16-12B-6*B^2+B^4
Input
Plots (figures that can be related to the open strings)
Alternate forms
128
Complex roots
Polynomial discriminant
Derivative
Local minimum
Definite integral
129
Definite integral area below the axis between the smallest and largest real
roots
6B^2+12B
Input
Plots (figures that can be related to the open strings)
130
Geometric figure
Alternate forms
Roots
Polynomial discriminant
Derivative
Indefinite integral
131
Global minimum
Definite integral
Definite integral area below the axis between the smallest and largest real
roots
From the ratio of the two previous expressions, we obtain also:
(16-12B-6*B^2+B^4) / (6B^2+12B)
Input
Plots (figures that can be related to the open strings)
132
Alternate forms
Expanded form
Quotient and remainder
133
Real roots
Series expansion at B=0
Series expansion at B=∞
Derivative
Indefinite integral
Local maximum
134
Local minima
Series representations
From:
we obtain:
(-16 - 16 B + 3 B^4 + B^5)/(3 B^2 (2 + B)^2)
Input
135
Plots (figures that can be related to the open strings)
Alternate forms
136
Expanded forms
Quotient and remainder
Real roots
137
Complex roots
Series expansion at B=0
Series expansion at B=∞
Derivative
Indefinite integral
Series representations
138
And again, from:
we obtain:
(B^6 + 6 B^5 + 12 B^4 + 48 B^2 + 96 B + 64)/(3 B^3 (2 + B)^3)
Input
Plots (figures that can be related to the open strings)
139
Alternate forms
Expanded forms
140
Complex roots
Series expansion at B=0
Series expansion at B=∞
141
Derivative
Indefinite integral
Local maximum
Local minimum
Limit
Series representations
142
From:
B^3/(3 B^3 + 18 B^2 + 36 B + 24) + (2 B^2)/(B^3 + 6 B^2 + 12 B + 8) + (4 B)/(B^3
+ 6 B^2 + 12 B + 8) + 16/((B^3 + 6 B^2 + 12 B + 8) B) + 32/((B^3 + 6 B^2 + 12 B +
8) B^2) + 64/(3 B^6 + 18 B^5 + 36 B^4 + 24 B^3)
(B^6 + 6 B^5 + 12 B^4 + 48 B^2 + 96 B + 64)/(3 B^3 (2 + B)^3)
For B = 4, we obtain:
64/(3*64 + 18*16 + 36*4 + 24) + 32/(64 + 6*16 + 48 + 8) + 16/(64 + 6*16 + 48 + 8)
+ 16/((64 + 6*16 + 48 + 8) 4) + 32/((64 + 6*16 + 48 + 8) 16) + 64/(3*4^6 + 18*4^5
+ 36*256 + 24*64)
Input
Exact result
143
Decimal approximation
0.3503086419753…..
From which:
sqrt(((0.9568666373/(((64/(64*3 + 18*16 + 144 + 24) + 32/(64 + 96 + 56) + 16/(64 +
96 + 56) + 16/((64 + 96 + 56) 4) + 32/((64 + 96 + 56) 16) + 64/(3*4^6 + 18*4^5 +
36*256 + 24*64)))))))
Input interpretation
Result
1.652723797678….. result very near to the 14th root of the following Ramanujan’s
class invariant =
505
/
101/5
3
= 1164.2696 i.e. 1.65578...
From:
for B = 4:
16-12*4-6*4^2+4^4
Input
144
Result
128
From which:
((27*1/2*(16-12*4-6*4^2+4^4)+1))
Input
Exact result
1729
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. (1728 = 8
2
* 3
3
) The number 1728 is one less than the Hardy–Ramanujan
number 1729 (taxicab number)
((27*1/2*(16-12*4-6*4^2+4^4)+1))^1/15
Input
Result
Decimal approximation
1.643815228…. ≈ ζ(2) =
2
6
= 1.644934 (trace of the instanton shape)
145
((1/2*(16-12*4-6*4^2+4^4)))^2
Input
Exact result
4096 = 64
2
From:
for B = 4
6*4^2+12*4
Input
Result
144
From which:
12(6*4^2+12*4)+1
Input
Result
1729
146
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. (1728 = 8
2
* 3
3
) The number 1728 is one less than the Hardy–Ramanujan
number 1729 (taxicab number)
((1/27*12(6*4^2+12*4)))^2
Input
Exact result
4096 = 64
2
((12(6*4^2+12*4)+1))^1/15
Input
Result
Decimal approximation
1.6438152287…. ≈ ζ(2) =
2
6
= 1.644934 (trace of the instanton shape)
147
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
148
“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
149
Appendix
From: A. Sagnotti – AstronomiAmo, 23.04.2020
In the above figure, it is said that: “why a given shape of the extra dimensions?
Crucial, it determines the predictions for α”.
We propose that whatever shape the compactified dimensions are, their geometry
must be based on the values of the golden ratio and ζ(2), (the latter connected to 1728
or 1729, whose fifteenth root provides an excellent approximation to the above
mentioned value) which are recurrent as solutions of the equations that we are going
to develop. It is important to specify that the initial conditions are always values
belonging to a fundamental chapter of the work of S. Ramanujan "Modular equations
and Appoximations to Pi" (see references). These values are some multiples of 8 (64
and 4096), 276, which added to 4096, is equal to 4372, and finally e
π√22
150
We have, in certain cases, the following connections:
Fig. 1
Fig. 2
151
Fig. 3
Stringscape - a small part of the string-theory landscape showing the new de Sitter solution as a local
minimum of the energy (vertical axis). The global minimum occurs at the infinite size of the extra
dimensions on the extreme right of the figure.
Fig. 4
152
With regard the Fig. 4 the points of arrival and departure on the right-hand side of the
picture are equally spaced and given by the following equation:
we obtain:
2Pi/(ln(2))
Input:
Exact result:
Decimal approximation:
9.06472028365….
Alternative representations:
153
Series representations:
Integral representations:
154
From which:
(2Pi/(ln(2)))*(1/12 π log(2))
Input:
Exact result:
Decimal approximation:
1.6449340668…. = ζ(2) =
2
6
= 1.644934
155
From:
Modular equations and approximations to - Srinivasa Ramanujan
Quarterly Journal of Mathematics, XLV, 1914, 350 – 372
We have that:
156
We note that, with regard 4372, we can to obtain the following results:
27((4372)^1/2-2-1/2(((√(10-2√5) -2))⁄((√5-1))))+φ
Input
Result
Decimal approximation
1729.0526944….
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. (1728 = 8
2
* 3
3
) The number 1728 is one less than the Hardy–Ramanujan
number 1729 (taxicab number)
Alternate forms
157
Minimal polynomial
Expanded forms
Series representations
158
Or:
27((4096+276)^1/2-2-1/2(((√(10-2√5) -2))⁄((√5-1))))+φ
159
Input
Result
Decimal approximation
1729.0526944…. as above
Alternate forms
160
Minimal polynomial
Expanded forms
Series representations
161
162
From which:
(27((4372)^1/2-2-1/2(((√(10-2√5) -2))⁄((√5-1))))+φ)^1/15
Input
Exact result
Decimal approximation
1.64381856858…. ≈ ζ(2) =
2
6
= 1.644934
Alternate forms
163
Minimal polynomial
Expanded forms
All 15th roots of ϕ + 27 (-2 + 2 sqrt(1093) - (sqrt(10 - 2 sqrt(5)) - 2)/(2 (sqrt(5) -
1)))
164
Series representations
165
Integral representation
166
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 64
2
, while -6C+ is equal to -
18. From this it follows that it is possible to establish mathematically, the dilaton
value.
167
For
exp((-Pi*sqrt(18)) we obtain:
Input:
Exact result:
Decimal approximation:
1.6272016… * 10
-6
Property:
Series representations:
168
Now, we have the following calculations:
6+
= 4096
18
18
= 1.6272016… * 10^-6
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...
169
Thence:
0.000244140625
6+
=
18
Dividing both sides by 0.000244140625, we obtain:
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:
170
Now:
6+
= 0.0066650177536
=
= 0.00666501785…
From:
ln(0.00666501784619)
Input interpretation:
Result:
-5.010882647757…
171
Alternative representations:
Series representations:
Integral representation:
In conclusion:
6 + = 5.010882647757
and for C = 1, we obtain:
172
= 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:
(http://www.bitman.name/math/article/102/109/)
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:
173
Result:
0.99040073.... result very near to the dilaton value . = and to
the value of the following Rogers-Ramanujan continued fraction:
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:
174
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
Input interpretation:
Result:
175
Solutions:
Alternate forms:
Expanded form:
Alternate form assuming b, M, and V are positive:
Alternate form assuming b, M, and V are real:
Derivative:
176
Implicit derivatives
Global minimum:
177
Global minima:
From:
we obtain
(225.913 (-0.054323 M^2 + 6.58545×10^-10 sqrt(M^4)))/M^2
Input interpretation:
Result:
178
Plots:
Alternate form assuming M is real:
-12.2723 result very near to the black hole entropy value 12.1904 = ln(196884)
Alternate forms:
179
Expanded form:
Property as a function:
Parity
Series expansion at M = 0:
Series expansion at M = ∞:
Derivative:
Indefinite integral:
180
Global maximum:
Global minimum:
Limit:
Definite integral after subtraction of diverging parts:
From b that is equal to
181
From:
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
Input interpretation:
Result:
Plots: (possible mathematical connection with an open string)
M = -0.5; M = 0.2
182
(possible mathematical connection with an open string)
M = 2 ; M = 3
Root:
Property as a function:
Parity
Series expansion at M = 0:
Series expansion at M = ∞:
183
Definite integral after subtraction of diverging parts:
For M = - 0.5 , we obtain:
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
184
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
Input interpretation:
Result:
7.021621519159*10
-17
For M = 3:
185
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
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
186
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:
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)))))
187
Input interpretation:
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
188
Planck’s Electric flux
5.975498*10
-8
V*m Lorentz-Heaviside value
Planck’s Electric potential
1.042940*10
27
V Lorentz-Heaviside value
189
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
190
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
191
References
Inflation after Planck and BICEP – Andrei Linde - “Quantum Gravity and All of
That” - https://qgholqi.inpcs.net/ - 13.01.2022 -
SHARP STABILITY INEQUALITIES FOR THE PLATEAU PROBLEM - G.
De Philippis & F. Maggi - j. differential geometry 96 (2014) 399-456
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
Properties of Nilpotent Supergravity
E. Dudas, S. Ferrara, A. Kehagias and A. Sagnotti - arXiv:1507.07842v2 [hep-th] 14
Sep 2015