1
On some equations concerning the general "unitary" formula, which derives
from DN Constant. New possible mathematical connections with the DN
Constant, Ramanujan Recurring Numbers and some parameters of Number
Theory, Cosmology and String Theory II
Michele Nardelli
1
, Antonio Nardelli
Abstract
In this paper (part II), we analyze some equations concerning the general "unitary"
formula, which derives from DN Constant. We obtain new possible mathematical
connections with the DN Constant, Ramanujan Recurring Numbers and some
parameters of Number Theory, Cosmology and String Theory
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
A. Nardelli studied at the Università degli Studi di Napoli Federico II - Dipartimento di Studi
Umanistici Sezione Filosofia - scholar of Theoretical Philosophy
2
Srinivasa Ramanujan (1887-1920)
https://www.moduscc.it/ramanujan-il-grande-matematico-indiano-13453-131115/
3
Introduction
In this paper, an octahedron could serve as a mathematical or conceptual model of the
universe in the quantic phase, while the spherical surface could be used to describe
the geometry of the bubble-universe.
The values (2√2)/π, the golden ratio φ, ζ(2) and π, can be connected to the proposed
cosmological model. Here's how they might be connected:
Ratio (2√2)/π the so called DN Constant:
This relationship may have a connection with the geometric properties of the
octahedron and the sphere, which have been considered as mathematical models of
the early universe and bubbles universe in eternal inflation.
Golden Ratio φ:
The golden ratio is a mathematical constant that appears in many natural and artistic
contexts and is often associated with harmonious proportions and aesthetic beauty. Its
emergence in this context could suggest a kind of intrinsic symmetry or harmony in
the structure of the early universe and bubbles universe.
Value of π:
The value of π is a fundamental mathematical constant that appears in many
geometric formulas and relationships, including the geometry of the sphere. Its
appearance could indicate a direct connection between the geometry of bubbles
universe and the mathematical properties of spherical surfaces.
Ultimately, the results obtained can be interpreted as manifestations of the geometric
and mathematical properties of the models proposed for the early universe and
universe bubbles. This suggests that there is a profound connection between
geometry, mathematics and cosmological physics, and that through the analysis of
4
these relationships we can deepen our understanding of the universe and its
fundamental phenomena.
The above values (2√2)/π, the golden ratio φ and π, can be connected to the proposed
cosmological model. This hypothesis is certainly plausible.
The various mathematical solutions and relationships can be seen as representations
of the principles and laws that govern the formation and evolution of the universe.
Regarding the fundamental mathematical values, they could emerge as a consequence
of the geometric and physical laws that govern the structure and evolution of the
quantum universe and bubbles universe.
The multidisciplinary approach involving complex mathematical solutions and
cosmological concepts can offer deeper insight into the fundamental nature of the
universe and its processes. Exploring these connections could lead to new discoveries
and insights into our understanding of the early universe and its complexity.
Proposal:
The initial octahedron: Let's imagine a regular octahedron, with perfectly
symmetrical faces. Each face represents an ideal symmetry.
The emerging sphere: Inside the octahedron, there is an inscribed sphere. This
sphere represents the bubble of the universe that emerges from the perturbations of
the quantum vacuum during eternal inflation.
Expansion and transitions: As time passes, the universe expands. The faces of the
octahedron begin to break, symbolizing "symmetry breaks." The sphere continues to
grow, representing the expanding universe.
Constants and numbers: We integrate the mathematical results you obtained. For
example, the golden ratio (φ) could be represented by a proportion between the
dimensions of the octahedron and the sphere.
Entropy and complexity: Entropy increases as the universe evolves. We can
represent this with a disordered growth of structures within the emerging sphere.
Imagine this scene as an abstract work of art, where geometric shapes and
cosmological concepts merge
5
In Fig.1 and Fig.2 let's imagine a regular octahedron representing the universe in its
phase of high symmetry and very low entropy. Inside the octahedron we have an
inscribed sphere that emerges from perturbations of the quantum vacuum during
eternal inflation. As time passes, the universe expands, the faces of the octahedron
break (symmetry breaks), and entropy increases. Spheres emerge from the octahedra,
symbolizing the transition phases from a regime of very low entropy to a universe in
which, with the passage of time, entropy increases, increasing the complexity of the
universe itself.
Fig. 1
6
Fig. 2
7
Now, we have that:
Octahedron Sphere
From the octahedron volume V = 1/3*√2 l
3
and, from the sphere volume,
V = (4/3*π*r
3
) , we consider the following relationship, for r = x:
4/3*π*x^3 = 1/3*√2*l^3
Input
Exact result
Alternate forms
8
Real solution
Solutions
Integer solution
Implicit derivatives
9
From the alternate form
for l = 8, we have that:
8/(sqrt(2) π^(1/3)) = 8/(2sqrt2 * Pi)^1/3
Input
Result
Logarithmic form
Thence:
l/(sqrt(2) π^(1/3)) = l/(2sqrt2 * Pi)^1/3
Input
10
Logarithmic form
Now, we have that:
l/(2 sqrt(2) π)^(1/3) = (2sqrt2)/Pi
Input
Exact result
Plot
11
Solution
12
From:
A NEW TYPE OF ISOTROPIC COSMOLOGICAL MODELS WITHOUT
SINGULARITY - A.A. STAROBINSKY - Volume 91B, number 1 PHYSICS
LETTERS 24 March 1980
We analyze the following equation:
We consider:
(d^2*f)/(d*ξ^2)+M^2*ξ^(-2/3)*f^(-1/3)-(M^2*f)/(12*H^2*ξ^2)+K(√12 M^2*f^(-
5/3)-3^(-1/2)*ξ^(-4/3)*f^(-1/3)*(M^2/H^2)+1)+K^2*ξ^(-2/3)*f^(-5/3)*(1-
(M^2/H^2))
Input
Exact result
Alternate forms
13
Derivative
Indefinite integral
14
From the indefinite integral result
we obtain:
(d^2 f)/(2 ξ^2) - (d f M^2)/(12 H^2 ξ^2) + (d (-K^2 M^2 + H^2 (K^2 + f^(4/3)
M^2)))/(f^(5/3) H^2 ξ^(2/3)) + d EllipticK(1 + (M^2 (6 - f^(4/3)/(H^2
ξ^(4/3))))/(sqrt(3) f^(5/3)))
Input
Exact result
Expanded form
15
Alternate forms
Derivative
Indefinite integral
16
Again, from the indefinite integral result
we obtain:
(d^3 f)/(6 ξ^2) - (d^2 f M^2)/(24 H^2 ξ^2) + (d^2 (-K^2 M^2 + H^2 (K^2 + f^(4/3)
M^2)))/(2 f^(5/3) H^2 ξ^(2/3)) + 1/2 d^2 EllipticK(1 + (M^2 (6 - f^(4/3)/(H^2
ξ^(4/3))))/(sqrt(3) f^(5/3)))
Input
Exact result
17
Alternate forms
Expanded form
Derivative
18
Indefinite integral
From the exact result
(d^3 f)/(6 ξ^2) - (d^2 f M^2)/(24 H^2 ξ^2) + (d^2 (-K^2 M^2 + H^2 (K^2 + f^(4/3)
M^2)))/(2 f^(5/3) H^2 ξ^(2/3)) + 1/2 d^2 EllipticK(1 + (M^2 (6 - f^(4/3)/(H^2
ξ^(4/3))))/(sqrt(3) f^(5/3)))
for d = 1, f = 2, ξ = 4, H = 8, K = 16:
(2)/(6 4^2) - (2 M^2)/(24 8^2 4^2) + ((-16^2 M^2 + 8^2 (16^2 + 2^(4/3) M^2)))/(2
2^(5/3) 8^2 4^(2/3)) + 1/2 EllipticK(1 + (M^2 (6 - 2^(4/3)/(8^2 4^(4/3))))/(sqrt(3)
2^(5/3)))
Input
19
Exact result
Plots (figures that can be related to the open strings)
Alternate forms
20
Expanded form
Numerical roots
21
Series expansion at M=0
Series expansion at M=∞
22
Derivative
Indefinite integral
23
From the exact result
1/48 - M^2/12288 + (-256 M^2 + 64 (256 + 2 2^(1/3) M^2))/1024 + 1/2 EllipticK(1 +
((6 - 1/(128 2^(1/3))) M^2)/(2 2^(2/3) sqrt(3)))
for M = 6:
1/48 - 6^2/12288 + (-256 6^2 + 64 (256 + 2 2^(1/3) 6^2))/1024 + 1/2 EllipticK(1 +
((6 - 1/(128 2^(1/3))) 6^2)/(2 2^(2/3) sqrt(3)))
Input
Exact result
Decimal approximation
Alternate complex forms
24
Alternate forms
Expanded forms
25
Alternative representations
Series representations
26
The general "unitary" formula, which derives from DN Constant, is the following:
 
 

 
    


  
for C = 1.616255*10
-35
(Planck Length), R = 1.265120782997423× 10
48
that is equal
to the radius of the Multiverse and 2.33*10
-13
is the mean temperature of the CMB is
equivalent around to 2.73 Kelvin , we obtain:
27
 
 

 


  



Multiplying the previous result 12.815 (cos(-0.019978) + i sin(-0.019978)) from the
above expression, we obtain:
(√(2×(2∙(((2√2)/π)^(1/16)))/(1/(π∙0.9991104684)(1.616255∙10^(-
35)×1.265120782997423*10^48×2.33∙10^(-13)))))(12.815(cos(-0.019978) + i sin(-
0.019978)))
Input interpretation
Result
Alternate complex forms
Polar coordinates
20.7351
28
From which, we obtain:
(76+7)*(√(2×(2∙(((2√2)/π)^(1/16)))/(1/(π∙0.9991104684)(1.616255∙10^(-
35)×1.265120782997423*10^48×2.33∙10^(-13)))))(12.815(cos(-0.019978) + i sin(-
0.019978)))+8
Input interpretation
Result
Alternate complex forms
Polar coordinates
1729.01
This result is very near to the mass of candidate glueball
 scalar meson.
Furthermore,  occurs in the algebraic formula for the j-invariant of an elliptic
curve 
 
. The number  is one less than the HardyRamanujan number
(taxicab number, as it can be expressed as the sum of two cubes in two different
ways 
 

 
 and Ramanujan's recurring number).
29
Since bosons are made of gauge bosons and scalar bosons (meson), then this number
theoretic analysis perhaps confirms that the number , confirm the fact that both
the gauge and scalar bosons are actually different states of a single bosonic string,
and that these states are isomorphic or that the states vibrations are synchronised with
the state of the bosonic string. This also imply that each state lives inside a cubic or
octahedron as a spherical cloud, and that the total sum of these two states is the state
of the bosonic string. Taking the cross section of the bosonic string, we realise that it
must be a rectangular, or a two shaped octahedron. As the string vibrates in
difference frequencies, so is the two spherical cloud states inside the string. That is,
the string vibrations simply excites the gauge bosons i.e Photon, gluon, W and Z
inside one cube/octahedron, and the scalar boson i.e. Higgs inside the other
cube/octahedron.
Furthermore, if we bring the picture of loop quantum gravity (LQG) with the
property of a discontinues quantum geometry, we can therefore, think of the graviton
living on the vertices of the rectangles or the octahedrons. This graviton then acts a
glue binding the bosonic strings lattice together forming a complete cross section of
alternating states of between the gauge bosons and scalar bosons. This arrangement
of states then gives a precise supersymmetric quantum picture of the vacuum
geometry at low entropy.
But the geometry further reveals very important fact, that since the vacuum geometry
is discontinues, then we observe that there is no relation whatsoever between the
quantum vibrational frequencies of the strings, and that of the vertices of the vacuum
geometry where the graviton lives. Ashtekar et al., (2021) asserted that gravity is
simply a manifestation of spacetime geometry. Thus, the graviton cannot be a string
boson, however, there is a duality between gravity and strings. Also, gauge bosons
have spin-1, while the graviton has spin-2. Then lastly, because of the
thermodynamic constraints we were able to arrive at the results we have, now this
bring us to this fundamental question; that string theory and LQG theory are two
intrinsic aspects of a complete quantum gravity theory we are after? That is, without
the other no complete and compelling quantum geometry can be attained, as it is done
here? This need to be investigated further.
30
(1/27(((76+7)*(√(2×(2∙(((2√2)/π)^(1/16)))/(1/(π∙0.9991104684)(1.616255∙10^(-
35)×1.265120782997423*10^48×2.33∙10^(-13)))))(12.815(cos(-0.019978) + i sin(-
0.019978)))+8)-1))^2
Input interpretation
Result
Alternate complex forms
Polar coordinates
4096.06 ≈ 4096
31
The number 4096 = 64
2
, is the Ramanujan Recurring Number, that when multiplied
by 2 give 8192. The total amplitude vanishes for gauge group SO(8192) for bosonic
string 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). This could be the implications for a pre-big bang scenario where only
self-perturbative bosonic strings lived when the enthalpy was extremely low as
discussed above. This regime contains all the intrinsic properties of superstrings
inherent in the bosonic strings, would at the big bang give effect to the properties of
matter (fermions) as Higgs Boson. This number theoretic connection to the gauge
group SO(8192), gives a much more compelling relevance of the bosonic string
theory SO(8192), to quantum gravity and places this string theory where it should
appropriately be in the evolution of the universe from a quantum gravity perspective
rather than it be neglected because it doesn’t include fermionic strings to confirm to
post big-bang reality. The vanishing of the bosonic string’s amplitude could be
explained by the effect of extreme low entropy on the quantum vacuum geometry.
Thus, as the entropy increases infinitesimally as a result of the vacuum self-
perturbation then also is the amplitude of the vibrating bosonic strings from zero.
Thus, was right to indicate that the “vanishing of the amplitude of the bosonic string
could be the results of string theory itself”, but here, we give a much more elaborate
explanation of what could be happening.
32
((76+7)*(√(2×(2∙(((2√2)/π)^(1/16)))/(1/(π∙0.9991104684)(1.616255∙10^(-
35)×1.265120782997423*10^48×2.33∙10^(-13)))))(12.815(cos(-0.019978) + i sin(-
0.019978)))+8)^1/15+(MRB const)^(1-1/(4π)+π)
Input interpretation
Result
Alternate complex forms
Polar coordinates
1.64494 ζ(2) = π
2
/6 = 1.644934 (trace of the instanton shape and Ramanujan
Recurring Number)
33
We analyze the following equation:
(k_2/(2880*π^2))(R^l*R-2/3*R*R_1-
1/2*g*R_2*R^(l*m)+1/4*g*R^2)+(k_3/(2880*π^2))*1/6*(2*R-2*g*R^l-
2*R*R_1+1/2*g*R^2)
Input
Exact result
Alternate forms
34
Expanded form
Periodicity
Root for the variable k
3
Series expansion at m=0
35
Derivative
Indefinite integral
From the indefinite integral result
we obtain:
(-(6 g k_2 R_2 R^(l m))/(l log(R)) + k_2 m R (3 g R + 12 R^l - 8 R_1) + k_3 m (R (g
R - 4 R_1 + 4) - 4 g R^l))/(34560 π^2)
Input
36
Expanded form
Alternate forms
Alternate forms assuming g, l, m, R, R
2
, R
1
, k
2
, and k
3
are positive
37
Reduced logarithmic form
Series expansion at m=0
Derivative
Indefinite integral
38
Again from the indefinite integral result
we obtain:
(-(12 g k_2 R_2 R^(l m))/(l^2 log^2(R)) + k_2 m^2 R (3 g R + 12 R^l - 8 R_1) + k_3
m^2 (R (g R - 4 R_1 + 4) - 4 g R^l))/(69120 π^2)
Input
Expanded form
Alternate forms
39
Alternate form assuming g, l, m, R, R
2
, R
1
, k
2
, and k
3
are positive
Reduced logarithmic form
Series expansion at m=0
Derivative
40
Indefinite integral
Again, from the indefinite integral result
(-(36 g k_2 R_2 R^(l m))/(l^3 log^3(R)) + k_2 m^3 R (3 g R + 12 R^l - 8 R_1) + k_3
m^3 (R (g R - 4 R_1 + 4) - 4 g R^l))/(207360 π^2)
for g = 2, k = 4, l = 8, m = 16 and R = 32:
(-(36 2 4 32 32^(128))/(8^3 log^3(32)) + 4 16^3 32 (3 2 32 + 12 32^8 - 8 32) + 4
16^3 (32 (2 32 - 4 32 + 4) - 4 2 32^8))/(207360 π^2)
Input
41
Decimal approximation
-9.6395909288989614…*10
185
Alternate forms
Expanded forms
42
Alternative representations
43
Series representations
44
Integral representations
45
From which, after some calculations:
4ln((-(36 2 4 32 32^(128))/(8^3 log^3(32)) + 4 16^3 32 (3 2 32 + 12 32^8 - 8 32) + 4
16^3 (32 (2 32 - 4 32 + 4) - 4 2 32^8))/(207360 π^2))+16
Input
Exact result
Decimal approximation
46
Alternate complex forms
47
Polar coordinates
1729
This result is very near to the mass of candidate glueball
 scalar meson.
Furthermore,  occurs in the algebraic formula for the j-invariant of an elliptic
curve 
 
. The number  is one less than the HardyRamanujan number
(taxicab number, as it can be expressed as the sum of two cubes in two different
ways 
 

 
 and Ramanujan's recurring number). Since bosons are
made of gauge bosons and scalar bosons (meson), then this number theoretic analysis
perhaps confirms that the number , confirm the fact that both the gauge and
scalar bosons are actually different states of a single bosonic string, and that these
states are isomorphic or that the states vibrations are synchronised with the state of
the bosonic string. This also imply that each state lives inside a cubic or octahedron
as a spherical cloud, and that the total sum of these two states is the state of the
bosonic string. Taking the cross section of the bosonic string, we realise that it must
be a rectangular, or a two shaped octahedron. As the string vibrates in difference
frequencies, so is the two spherical cloud states inside the string.
48
That is, the string vibrations simply excites the gauge bosons i.e Photon, gluon, W
and Z inside one cube/octahedron, and the scalar boson i.e. Higgs inside the other
cube/octahedron.
Furthermore, if we bring the picture of loop quantum gravity (LQG) with the
property of a discontinues quantum geometry, we can therefore, think of the graviton
living on the vertices of the rectangles or the octahedrons. This graviton then acts a
glue binding the bosonic strings lattice together forming a complete cross section of
alternating states of between the gauge bosons and scalar bosons. This arrangement
of states then gives a precise supersymmetric quantum picture of the vacuum
geometry at low entropy.
But the geometry further reveals very important fact, that since the vacuum geometry
is discontinues, then we observe that there is no relation whatsoever between the
quantum vibrational frequencies of the strings, and that of the vertices of the vacuum
geometry where the graviton lives. Ashtekar et al., (2021) asserted that gravity is
simply a manifestation of spacetime geometry. Thus, the graviton cannot be a string
boson, however, there is a duality between gravity and strings. Also, gauge bosons
have spin-1, while the graviton has spin-2. Then lastly, because of the
thermodynamic constraints we were able to arrive at the results we have, now this
bring us to this fundamental question; that string theory and LQG theory are two
intrinsic aspects of a complete quantum gravity theory we are after? That is, without
the other no complete and compelling quantum geometry can be attained, as it is done
here? This need to be investigated further.
Expanded forms
49
Alternate forms
50
Alternative representations
51
(1/27((4ln((-(36 2 4 32 32^(128))/(8^3 log^3(32)) + 4 16^3 32 (3 2 32 + 12 32^8 - 8
32) + 4 16^3 (32 (2 32 - 4 32 + 4) - 4 2 32^8))/(207360 π^2))+16)-1))^2
Input
Decimal approximation
Alternate complex forms
52
53
Polar coordinates
4096.1 ≈ 4096
The number 4096 = 64
2
, is the Ramanujan Recurring Number, that when multiplied
by 2 give 8192. The total amplitude vanishes for gauge group SO(8192) for bosonic
string SO(8192), while the vacuum energy is negative and independent of the gauge
group.
54
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). This could be the implications for a pre-big bang scenario where only self-
perturbative bosonic strings lived when the enthalpy was extremely low as discussed
above. This regime contains all the intrinsic properties of superstrings inherent in the
bosonic strings, would at the big bang give effect to the properties of matter
(fermions) as Higgs Boson. This number theoretic connection to the gauge group
SO(8192), gives a much more compelling relevance of the bosonic string theory
SO(8192), to quantum gravity and places this string theory where it should
appropriately be in the evolution of the universe from a quantum gravity perspective
rather than it be neglected because it doesn’t include fermionic strings to confirm to
post big-bang reality. The vanishing of the bosonic string’s amplitude could be
explained by the effect of extreme low entropy on the quantum vacuum geometry.
Thus, as the entropy increases infinitesimally as a result of the vacuum self-
perturbation then also is the amplitude of the vibrating bosonic strings from zero.
Thus, was right to indicate that the “vanishing of the amplitude of the bosonic string
could be the results of string theory itself”, but here, we give a much more elaborate
explanation of what could be happening.
(4ln((-(36 2 4 32 32^(128))/(8^3 log^3(32)) + 4 16^3 32 (3 2 32 + 12 32^8 - 8 32) + 4
16^3 (32 (2 32 - 4 32 + 4) - 4 2 32^8))/(207360 π^2))+16)^1/15+(MRB const)^(1-
1/(4π)+π)
Input
55
Exact result
Decimal approximation
Alternate complex forms
Polar coordinates
1.6449 ζ(2) = π
2
/6 = 1.644934 (trace of the instanton shape and Ramanujan
Recurring Number)
56
Alternate forms
The general "unitary" formula, which derives from DN Constant, is the following:
 
 

 
    


  
57
Where
is the Del Gaudio-Nardelli Constant, 0.9991104684 is the value of the
following Rogers-Ramanujan continued fraction:
C is any constant or solution to an equation, R is the radius of the Universe and
2.33*10
-13
is the temperature of the universe expressed in GeV.
And, always from:
 
 

 


   

for C = 1.616255*10
-35
(Planck Length), R = 1.265120782997423× 10
48
that is equal
to the radius of the Multiverse and 2.33*10
-13
is the mean temperature of the CMB is
equivalent around to 2.73 Kelvin, multiplying
 
 

 


  



58
by the previous expression
we obtain:
(√(2×(2∙(((2√2)/π)^(1/16)))/(1/(π∙0.9991104684)(1.616255∙10^(-
35)×1.265120782997423*10^48×2.33∙10^(-13))))) (((-(36 2 4 32 32^(128))/(8^3
log^3(32)) + 4 16^3 32 (3 2 32 + 12 32^8 - 8 32) + 4 16^3 (32 (2 32 - 4 32 + 4) - 4 2
32^8))/(207360 π^2)))
Input interpretation
Result
-1.55972…*10
186
59
From which, after some calculations:
4ln((√(2×(2∙(((2√2)/π)^(1/16)))/(1/(π∙0.9991104)(1.616255e-
35×1.2651207829e+48×2.33e-13))))(((-(36*256 32^128)/(8^3 log^3(32))+16^3
128(192+12 32^8-256)+4 16^3 (32(-64+4)-8 32^8))/(207360π^2))))+11+3
Input interpretation
Result
Alternate complex forms
Polar coordinates
1728.95
This result is very near to the mass of candidate glueball
 scalar meson.
Furthermore,  occurs in the algebraic formula for the j-invariant of an elliptic
curve 
 
. The number  is one less than the HardyRamanujan number
(taxicab number, as it can be expressed as the sum of two cubes in two different
ways 
 

 
 and Ramanujan's recurring number).
60
Since bosons are made of gauge bosons and scalar bosons (meson), then this number
theoretic analysis perhaps confirms that the number , confirm the fact that both
the gauge and scalar bosons are actually different states of a single bosonic string,
and that these states are isomorphic or that the states vibrations are synchronised with
the state of the bosonic string. This also imply that each state lives inside a cubic or
octahedron as a spherical cloud, and that the total sum of these two states is the state
of the bosonic string. Taking the cross section of the bosonic string, we realise that it
must be a rectangular, or a two shaped octahedron. As the string vibrates in
difference frequencies, so is the two spherical cloud states inside the string. That is,
the string vibrations simply excites the gauge bosons i.e Photon, gluon, W and Z
inside one cube/octahedron, and the scalar boson i.e. Higgs inside the other
cube/octahedron.
Furthermore, if we bring the picture of loop quantum gravity (LQG) with the
property of a discontinues quantum geometry, we can therefore, think of the graviton
living on the vertices of the rectangles or the octahedrons. This graviton then acts a
glue binding the bosonic strings lattice together forming a complete cross section of
alternating states of between the gauge bosons and scalar bosons. This arrangement
of states then gives a precise supersymmetric quantum picture of the vacuum
geometry at low entropy.
But the geometry further reveals very important fact, that since the vacuum geometry
is discontinues, then we observe that there is no relation whatsoever between the
quantum vibrational frequencies of the strings, and that of the vertices of the vacuum
geometry where the graviton lives. Ashtekar et al., (2021) asserted that gravity is
simply a manifestation of spacetime geometry. Thus, the graviton cannot be a string
boson, however, there is a duality between gravity and strings. Also, gauge bosons
have spin-1, while the graviton has spin-2. Then lastly, because of the
thermodynamic constraints we were able to arrive at the results we have, now this
bring us to this fundamental question; that string theory and LQG theory are two
intrinsic aspects of a complete quantum gravity theory we are after? That is, without
the other no complete and compelling quantum geometry can be attained, as it is done
here? This need to be investigated further.
61
(1/27((4ln((√(2(2(((2√2)/π)^(1/16)))/(1/(π0.9991104)(1.61625e-
35×1.26512e+48×2.33e-13))))(((-(36*256 32^128)/(512ln^3(32))+16^3 128(192+12
32^8-256)+4 16^3(-1920-8 32^8))/(207360π^2))))+14)-1))^2+1/4
Input interpretation
Result
Alternate complex forms
Polar coordinates
4096
The number 4096 = 64
2
, is the Ramanujan Recurring Number, that when multiplied
by 2 give 8192. The total amplitude vanishes for gauge group SO(8192) for bosonic
string 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.
62
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). This could
be the implications for a pre-big bang scenario where only self-perturbative bosonic
strings lived when the enthalpy was extremely low as discussed above. This regime
contains all the intrinsic properties of superstrings inherent in the bosonic strings,
would at the big bang give effect to the properties of matter (fermions) as Higgs
Boson. This number theoretic connection to the gauge group SO(8192), gives a much
more compelling relevance of the bosonic string theory SO(8192), to quantum
gravity and places this string theory where it should appropriately be in the evolution
of the universe from a quantum gravity perspective rather than it be neglected
because it doesn’t include fermionic strings to confirm to post big-bang reality. The
vanishing of the bosonic string’s amplitude could be explained by the effect of
extreme low entropy on the quantum vacuum geometry. Thus, as the entropy
increases infinitesimally as a result of the vacuum self-perturbation then also is the
amplitude of the vibrating bosonic strings from zero. Thus, was right to indicate that
the “vanishing of the amplitude of the bosonic string could be the results of string
theory itself”, but here, we give a much more elaborate explanation of what could be
happening.
(4ln((√(2×(2∙(((2√2)/π)^(1/16)))/(1/(π∙0.9991104)(1.616255e-
35×1.26512078e+48×2.33e-13))))(((-(36*256 32^128)/(8^3 ln^3(32))+16^3
128(192+12 32^8-256)+4 16^3 (-1920-8 32^8))/(207360π^2))))+11+3)^1/15
Input interpretation
63
Result
Alternate complex forms
Polar coordinates
1.64381
((1.64381 (cos(0.000484552) + i sin(0.000484552))))+(MRB const)^(1-1/(4π)+π)
Input interpretation
Result
Alternate complex forms
Polar coordinates
1.64493 ζ(2) = π
2
/6 = 1.644934 (trace of the instanton shape and Ramanujan
Recurring Number)
64
Now, we analyze the following equation:
we consider:
(δ’’*(1-K/(H^2*a^2))+δ’/a*(4-(5K/(H^2*a^2)))-((M^2*δ)/(H^2*a^2)))
Input
Alternate forms
Expanded form
65
Derivative
From the expanded form
-(5 K δ'(M))/(a^3 H^2) - (K δ''(M))/(a^2 H^2) - (M^2 δ(M))/(a^2 H^2) + (4 δ'(M))/a
+ δ''(M)
we know that:
thus:
-(5 K)/(a^3 H^2) - (K*0)/(a^2 H^2) - (M^2 δ(M))/(a^2 H^2) + (4)/a + 0
Input
Result
66
Alternate forms
Partial fraction expansion
The general "unitary" formula, which derives from DN Constant, is the following:
 
 

 
    


  
for C = 1.616255*10
-35
(Planck Length), R = 1.265120782997423× 10
48
that is equal
to the radius of the Multiverse and 2.33*10
-13
is the mean temperature of the CMB is
equivalent around to 2.73 Kelvin , we obtain:
67
 
 

 
 

 

  


multiplying
√(2×(2∙(((2√2)/π)^(1/16)))/(1/(π∙0.9991104684)(1.616255∙10^(-
35)×1.265120782997423*10^48×2.33∙10^(-13))))
by the previous result
-(5 K)/(a^3 H^2) - (M^2 δ(M))/(a^2 H^2) + 4/a
we obtain:
(√(2×(2∙(((2√2)/π)^(1/16)))/(1/(π∙0.9991104684)(1.616255∙10^(-
35)×1.265120782997423*10^48×2.33∙10^(-13)))))*((-(5 K)/(a^3 H^2) - (M^2
δ(M))/(a^2 H^2) + 4/a))
Input interpretation
Result
68
Alternate forms
Expanded form
Partial fraction expansion
From the result
1.61803 (-(5 K)/(a^3 H^2) - (M^2 δ(M))/(a^2 H^2) + 4/a)
for
; k
3
= - 2 k
2
= 2 ; K = 1
69
M
2
= ((-360π)/(-2*G))
H
2
= ((360π)/(2*G))
we obtain:
1.61803 (-(5)/(a^3 (((360π)/(2*G)))) - ((((-360π)/(-2*G))) δ(((-360π)/(-
2*G))^1/2))/(a^2 (((360π)/(2*G)))) + 4/a)
Input interpretation
Result
70
3D plots
Real part (figures that can be related to the D-branes/Instantons)
The key observation from the below plots and is that at , which is taken as the
energy density of the universe at the Big Bang, with the zero spacetime volume,
the vacuum geometry brakes / or there is symmetry breaking on the vacuum quantum
geometry. We see from the plots as the vacuum spacetime break/tear apart.
Imaginary part
71
Contour plots
Real part
Imaginary part
Partial fraction expansion
72
Alternate forms assuming a and G are positive
Alternate form assuming a and G are real
Alternate forms
From the result
1.61803 (-G/(36 π a^3) - δ(sqrt(1/G))/(6 sqrt(5 π) a^2) + 4/a)
for a = 0.5 and G = 1:
73
1.61803 (-1/(36 π 0.5^3) - δ(sqrt(1/1))/(6 sqrt(5 π) 0.5^2) + 4/0.5)
Input interpretation
Result
12.8298
From which:
((1/(2Pi))(1/4((1.61803 (-1/(36 π 0.5^3) - δ(sqrt(1/1))/(6 sqrt(5 π) 0.5^2) +
4/0.5))))^2)
Input interpretation
Result
1.6374…. result quite near to the mean between ζ(2) =
, the value of
golden ratio 1.61803398… and the 14th root of the Ramanujan’s class invariant



= 1164.2696 i.e. 1.65578..., i.e. 1.63958266
74
1+1/(((1.61803 (-1/(36 π 0.5^3) - δ(sqrt(1/1))/(6 sqrt(5 π) 0.5^2) + 4/0.5)))^1/6)
Input interpretation
Result
1.65358 result very near to the 14th root of the following Ramanujan’s class
invariant 


= 1164.2696 i.e. 1.6557845...
From the mean of the two expressions, we obtain:
1/2((1/(2Pi))(1/4((1.61803 (-1/(36 π 0.5^3) - δ(sqrt(1/1))/(6 sqrt(5 π) 0.5^2) +
4/0.5))))^2+(1+1/(((1.61803 (-1/(36 π 0.5^3) - δ(sqrt(1/1))/(6 sqrt(5 π) 0.5^2) +
4/0.5)))^1/6)))
Input interpretation
75
Result
1.64546 ζ(2) = π
2
/6 = 1.644934 (trace of the instanton shape and Ramanujan
Recurring Number)
Now, we analyze the following equation:
(M^2*(x-x^(-1/3)))/((12*H^2*y)-1)
Input
Result
Expanded form
Alternate form assuming H, M, x, and y are positive
76
Alternate form
Roots
Properties as a function
Domain
Range
Series expansion at x=0
Series expansion at x=∞
77
Derivative
Indefinite integral
Limit
Series representations
78
From the indefinite integral result
we obtain:
(M^2 (-3 x^(2/3) + x^2))/(-2 + 24 H^2 y)
Input
Result
Expanded form
Alternate form assuming H, M, x, and y are positive
Alternate form
79
Roots
Properties as a function
Domain
Range
Series expansion at x=0
Series expansion at x=∞
80
Derivative
Indefinite integral
Limit
Series representations
81
Again, from the indefinite integral result
we obtain:
(M^2 (-27 x^(5/3) + 5 x^3))/(-30 + 360 H^2 y)
Input
Result
Expanded form
Alternate form assuming H, M, x, and y are positive
Alternate form
82
Roots
Properties as a function
Domain
Range
Series expansion at x=0
Series expansion at x=∞
83
Derivative
Indefinite integral
Limit
Series representations
84
From the indefinite integral result
(M^2 (-81 x^(8/3) + 10 x^4))/(240 (-1 + 12 H^2 y))
The general "unitary" formula, which derives from DN Constant, is the following:
 
 

 
    


  
for C = 1.616255*10
-35
(Planck Length), R = 1.265120782997423× 10
48
that is equal
to the radius of the Multiverse and 2.33*10
-13
is the mean temperature of the CMB is
equivalent around to 2.73 Kelvin , we obtain:
 
 

 
 

 

  


multiplying
√(2×(2∙(((2√2)/π)^(1/16)))/(1/(π∙0.9991104684)(1.616255∙10^(-
35)×1.265120782997423*10^48×2.33∙10^(-13))))
by the previous result
(M^2 (-81 x^(8/3) + 10 x^4))/(240 (-1 + 12 H^2 y))
85
we obtain:
((√(2×(2∙(((2√2)/π)^(1/16)))/(1/(π∙0.9991104684)(1.616255∙10^(-
35)×1.265120782997423*10^48×2.33∙10^(-13))))))*(((M^2 (-81 x^(8/3) + 10
x^4))/(240 (-1 + 12 H^2 y))))
Input interpretation
Result
Alternate form assuming H, M, x, and y are real
Expanded form
Alternate forms
86
Roots
Properties as a function
Domain
Range
Series expansion at x=0
87
Series expansion at x=∞
Derivative
Indefinite integral
Limit
From the indefinite integral result
(M^2 (-0.0124111 x^(11/3) + 0.00112363 x^5))/(-0.0833333 + 1 H^2 y)
88
for x = 2, y = 4, H = 8 and M = 16:
(16^2 (-0.0124111 2^(11/3) + 0.00112363 2^5))/(-0.0833333 + 1 8^2 4)
Input interpretation
Result
-0.12169459968….
From which, after some calculations:
1+((2 (3/5)^(2/3))/e^(1/3))*1/((-13((16^2 (-0.0124111 2^(11/3) + 0.00112363
2^5))*1/(-0.0833333 + 1 8^2 4))))
where
and the inverse is 0.9809209886.
With regard this value, we note 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. (Astronomy
& Astrophysics manuscript no. ms c ESO 2019 - September 24, 2019 - Planck 2018
results. VI. Cosmological parameters). Furthermore there is also the following
Scalar Spectral Index value (n
s
= 0.957-0.976)
89
Input interpretation
Result
1.64439378518…. ζ(2) = π
2
/6 = 1.644934 (trace of the instanton shape and
Ramanujan Recurring Number)
Alternative representation
Series representations
90
sqrt(6*(1+((2 (3/5)^(2/3))/e^(1/3))*1/((-13((16^2 (-0.0124111 2^(11/3) + 0.00112363
2^5))*1/(-0.0833333 + 1 8^2 4))))))
Input interpretation
Result
3.1410766802313…. ≈ π (Ramanujan Recurring Number)
All 2
nd
roots of 9.86636
Series representations
91
From the following formula
(4 (1.6180085459)^(2 π))/(5 (3 + sqrt(5))) = 3.1415926535
we obtain:
(4 (x)^(2 π))/(5 (3 + sqrt(5))) = 3.14107668
Input interpretation
92
Results
Plot (figure that can be related to an open string)
Alternate form assuming x is real
Alternate forms
Real solution
1.61797 result that is a very good approximation to the value of the golden ratio
1.618033988749… (Ramanujan Recurring Number)
93
In the below figures, Multiverse eternal inflation with cosmic bubbles inscribed in
octahedral shapes
Fig. 3
94
Fig. 4
Fig. 5
Fig. 6
95
On the application of the formulas of the volumes of an octahedron and a sphere
With regard to a sphere inscribed in an octahedron, we have the following formulas.
Fig: sphere inscribed in an octahedron
V
0
=

V
s
=
 where r
s
= (l/2)
With regard the ratio between the two above formulas (octahedron and sphere)
(1/3*√2*l^3)/(4/3*π*(l/2)^3)
we obtain:
Input
96
Result
Decimal approximation
0.900316316157106…. =
(DN Constant)
Property
Series representations
97
From which:
1/3*(2/((1/3*√2*l^3)/(4/3*π*(l/2)^3)))^2
Input
Result
Decimal approximation
1.644934066848226… = ζ(2) = π
2
/6 = 1.644934 (trace of the instanton shape and
Ramanujan Recurring Number)
98
Property
Series representations
Integral representations
99
We note that, from the sum of the first nine numbers excluding 0, i.e.,
1+2+3+4+5+6+7+8+9 = 45 (these are the fundamental numbers, from which, through
infinite combinations, all the other numbers are obtained), we obtain the following
interesting formula:
1+1/(((φ^2+(2Pi)/3*MRB const)(1/e((1+2+3+4+5+6+7+8+9)^(1/Pi))))^1/3)
Input
Exact result
100
Decimal approximation
1.64529737852…. ≈ ζ(2) = π
2
/6 = 1.644934 (trace of the instanton shape and
Ramanujan Recurring Number)
Alternate forms
Expanded forms
101
And:
sqrt(6(1+1/(((φ^2+(2Pi)/3*MRB const)(1/e((1+2+3+4+5+6+7+8+9)^(1/Pi))))^1/3)))
Input
Exact result
Decimal approximation
3.141939571526…. ≈ π (Ramanujan Recurring Number)
Alternate forms
102
Expanded forms
All 2
nd
roots of 6 (3
-2/(3 π)
5
-1/(3 π)
(e/((2 π C
MRB
)/3+ϕ
2
))
1/3
+1)
Furthermore, we obtain also:
2π*√2((1/3*√2*l^3)/(4/3*π*(l/2)^3))
Input
103
Exact result
8
value that is linked to the "Ramanujan function" (an elliptic modular function that
satisfies the need for "conformal symmetry") that has 8 "modes" corresponding to the
physical vibrations of a superstring.
Series representations
104
6π*√2((1/3*√2*l^3)/(4/3*π*(l/2)^3))
Input
Exact result
24
The value 24 is linked to the "Ramanujan function" (an elliptic modular function that
satisfies the need for "conformal symmetry") that has 24 "modes" corresponding to
the physical vibrations of a bosonic string representing a bosons. From the analysis,
we observe that the is no number theoretic connection with physical vibrations of
fermionic strings at extremally low entropy. This fact is confirmed by the fact that the
Higgs bosons at the moment of the big bang and infinitesimally shortly thereafter,
facilitated the creation of fermions (matter and antimatter particles) [8]. Thus we note
that the ingredients for the formation of electromagnetic radiation from photons (a
Boson), and the formation of matter from the Higgs boson after the big bang, are
intrinsic properties of the vacuum energy in pre-big bang.
Series representations
105
This could imply that all matter (fermions) was preceded by bosons. That is, before
the Big Bang, from perturbations of the vacuum energy itself, bosons were created,
and successively at the Big Bang, and infinitesimally shortly after the Big Bang,
fermions, were created from the vacuum energy that underwent a violent “breaking”
that formed a hot plasma. of particle-antiparticle pairs. This therefore implies that
quantum gravity was not necessarily “dark” to some extent, because a photon (light
particle) is itself a boson. Therefore, a big bang was not necessarily the moment of
the creation of light, but of the creation of matter (fermions) from vacuum energy, as
this undergoes further "breaking" in the cosmological constant, in the hot plasma of
matter and in the energy dark.
106
(2π*√2((1/3*√2*l^3)/(4/3*π*(l/2)^3)))^4
Input
Exact result
4096 = 64
2
, (Ramanujan Recurring Number) that multiplied by 2 give 8192, indeed:
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)
27*sqrt((2π*√2((1/3*√2*l^3)/(4/3*π*(l/2)^3)))^4)+1
Input
Exact result
1729
107
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 HardyRamanujan
number 1729 (taxicab number, as it can be expressed as the sum of two cubes in two
different ways (10
3
+ 9
3
= 12
3
+ 1
3
= 1729) and Ramanujan's recurring number)
Series representations
108
We note that:
1/25*1/144(((2π*√2((1/3*√2*l^3)/(4/3*π*(l/2)^3)))^4)+(27*sqrt((2π*√2((1/3*√2*l^
3)/(4/3*π*(l/2)^3)))^4)+1))
Input
Exact result
Decimal approximation
1.61805555…. result that is a very good approximation to the value of the golden
ratio 1.618033988749… (Ramanujan Recurring Number)
Repeating decimal
109
Series representations
110
From
we obtain also:
sqrt(6(1/3*(2/(((2sqrt2)/Pi)))^2))
Input
Exact result
Decimal approximation
3.14159265358… = π
111
Property
All 2
nd
roots of π
2
Series representations
Integral representations
112
It is plausible to hypothesize that π and φ, in addition to being important
mathematical constants, are constants that also have a fundamental relevance in the
various sectors of Theoretical Physics and Cosmology
From
, we obtain:
sqrt(1/(Pi^2/6)*(4/3))
Input
Exact result
113
Decimal approximation
0.900316316157106…. =
(DN Constant)
Property
All 2
nd
roots of 8/π
2
Series representations
114
DN Constant extended
We have the following expression concerning the ratios (and/or the inverses)
between the icosahedron, octahedron and tetrahedron volumes and the sphere
volume.

 



(we have highlighted the DN Constant in blue)
(((((5/12*(3+√5)*d^3)/(4/3*π*(d/2)^3))*1/((1/3*√2*a^3)/(4/3*π*(a/2)^3)) *1/((((√2
d^3)/12))*1/(4/3*π*(d/2)^3)))))^(1/(2π))
115
Input
Exact result
Decimal approximation
1.6180085459…. result that is a very good approximation to the value of the golden
ratio 1.618033988749… (Ramanujan Recurring Number)
Alternate form
Series representations
116
117
Integral representation
Furthermore, from the formula
we obtain also:
(4 (1.6180085459)^(2 π))/(5 (3 + sqrt(5))) = 3.1415926535
Indeed:
3.1415926535…. = π (Ramanujan Recurring Number)
118
From the following extended DN Constant (“Unitary Formula”)

 





with regard


for q = 1729 and p = 4096, we obtain by changing the sign in the algebraic sum of
the aforementioned Cardano’s Formula and after some calculations:
multiplied by
and performing the ninth root of the entire expression:
√((2^(-1/π)(5(3+√(5))π)^(1/(2π)))(((-1729/2+√(1729^2/4+4096^3/27)) -(-1729/2-
√(1729^2/4+4096^3/27)))^1/9))
119
we obtain:
i.e.
2^(-1/(2π)) ((-1729/2+√(274958621851/3)/6)^(1/3)+(1729/2 +
√(274958621851/3)/6)^(1/3))^(1/18) (5(3+√(5)) π)^(1/(4π))
= 1.61549140391….
The general "unitary" formula, which derives from DN Constant, is the following:
 
 

  
   


  
120
Where
is the Del Gaudio-Nardelli Constant, 0.9991104684 is the value of the
following Rogers-Ramanujan continued fraction:
C is any constant or solution to an equation, R is the radius of the Universe and
2.33*10
-13
is the temperature of the universe expressed in GeV.
For example, C = 9.9128
, inserting a radius of the Universe, which we have
calculated, equal to R = 2.06274*10
12
years, from DN Constant "unitary" formula, we
obtain:
√(2×(2∙(((2√2)/π))^(1/16))/(1/(π∙0.9991104684) (9.9128×(2.06274 ×
10^12)×2.33∙10^(-13))))
1.618035912348…. result that is a very good approximation to the value of the
golden ratio 1.618033988749… (Ramanujan Recurring Number)
121
We obtain also:
(√(2×(2∙(((2√2)/π))^(1/16))/(1/(π∙0.9991104684) (9.9128×(2.06274 ×
10^12)×2.33∙10^(-13)))))dxdydz
Indefinite integral assuming all variables are real
Definite integral over a cube of edge length 2 L
Definite integral over a sphere of radius R
From which, for L = R = 1 , dividing the two definite integral results by the original
expression, we obtain:
12.9443/(√(2×(2∙(((2√2)/π))^(1/16))/(1/(π∙0.9991104684) (9.9128×(2.06274 ×
10^12)×2.33∙10^(-13)))))
Input interpretation
122
Result
8.00001…. ≈ 8
value that is linked to the "Ramanujan function" (an elliptic modular function that
satisfies the need for "conformal symmetry") that has 8 "modes" corresponding to the
physical vibrations of a superstring.
And
3*(6.77761/(√(2×(2∙(((2√2)/π))^(1/16))/(1/(π∙0.9991104684) (9.9128×(2.06274 ×
10^12)×2.33∙10^(-13))))))
Input interpretation
Result
12.5664…. ≈ 4π = Bekenstein-Hawking (S
BH
) black hole entropy
123
Now, we have that:
Octahedron Sphere
Given the value of a volume, independently of the solid, following the Poincaré
Conjecture, we compare any solid "without holes" and a sphere. If we compare an
octahedron with a sphere, we have:

If we consider the radius of the sphere as an unknown, we must find the value of
the side of the octahedron which allows us to equalize the two volumes and which
will give us the DN Constant as a result (which will therefore be equal to the
radius of the sphere).
From

To find we perform the following calculation:





 

  

  

124




To find a, we have, for



Thus, multiplying both the sides by

, we obtain:


  
 
    
 

  

  





125
Plot
Solution
for V = 1/3*√2*a^3 (octahedron volume) and V = (4/3*π*r^3) (sphere volume), we
obtain:
from the octahedron volume, we have: V = 1/3*√2*a
3
= (1/3*√2*(
)^3)
(1/3*√2*(4/(π^2 ))^3)
Input
Exact result
126
Decimal approximation
3.05684889733….
Property
Series representations
127
And, from the sphere volume V = (4/3*π*r
3
) = (4/3*π*((2√2)/π)^3)
(4/3*π*((2√2)/π)^3)
Input
Result
Decimal approximation
3.05684889733….
Property
Series representations
128
129
From:
Modular equations and approximations to - Srinivasa Ramanujan - Quarterly
Journal of Mathematics, XLV, 1914, 350 372
We have that:
130
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 HardyRamanujan
number 1729 (taxicab number)
Alternate forms
131
Minimal polynomial
Expanded forms
Series representations
132
133
Or:
27((4096+276)^1/2-2-1/2(((√(10-2√5) -2))⁄((√5-1))))
Input
Result
Decimal approximation
1729.0526944…. as above
Alternate forms
134
Minimal polynomial
Expanded forms
Series representations
135
136
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) =

Alternate forms
137
Minimal polynomial
Expanded forms
All 15th roots of ϕ + 27 (-2 + 2 sqrt(1093) - (sqrt(10 - 2 sqrt(5)) - 2)/(2 (sqrt(5) -
1)))
138
Series representations
139
Integral representation
140
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:




141
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 -
. From this it follows that it is possible to establish mathematically, the dilaton
value.
For
exp((-Pi*sqrt(18)) we obtain:
Input:
Exact result:
Decimal approximation:
1.6272016… * 10
-6
Property:
Series representations:
142
Now, we have the following calculations:






= 1.6272016… * 10
-6
from which:


= 1.6272016… * 10
-6
0.000244140625

=


= 1.6272016… * 10
-6
Now:





And:
143
(1.6272016* 10^-6) *1/ (0.000244140625)
Input interpretation:
Result:
0.006665017...
Thence:
0.000244140625

=


Dividing both sides by 0.000244140625, we obtain:



=




= 0.0066650177536
((((exp((-Pi*sqrt(18)))))))*1/0.000244140625
Input interpretation:
144
Result:
0.00666501785…
Series representations:
Now:

= 0.0066650177536
=
= 0.00666501785…
145
From:
ln(0.00666501784619)
Input interpretation:
Result:
-5.010882647757…
Alternative representations:
Series representations:
146
Integral representation:
In conclusion:
 
and for C = 1, we obtain:
   =
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 , are also connected to
the following two Rogers-Ramanujan continued fractions:
147
(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:
Result:
0.99040073.... result very near to the dilaton value  = and to the
value of the following Rogers-Ramanujan continued fraction:
148
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:
We have:
149
(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:
150
Solutions:
Alternate forms:
Expanded form:
Alternate form assuming b, M, and V are positive:
Alternate form assuming b, M, and V are real:
Derivative:
151
Implicit derivatives
Global minimum:
Global minima:
152
From:
we obtain:
(225.913 (-0.054323 M^2 + 6.58545×10^-10 sqrt(M^4)))/M^2
Input interpretation:
Result:
153
Plots:
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:
154
Property as a function:
Parity
Series expansion at M = 0:
Series expansion at M = ∞:
Derivative:
Indefinite integral:
155
Global maximum:
Global minimum:
Limit:
Definite integral after subtraction of diverging parts:
From b that is equal to
156
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
(possible mathematical connection with an open string)
M = 2 ; M = 3
157
Root:
Property as a function:
Parity
Series expansion at M = 0:
Series expansion at M = ∞:
Definite integral after subtraction of diverging parts:
158
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
For M = 0.2:
159
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:
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
160
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:
sqrt[1/(2Pi)(7.021621519*10^-15 + 1.57986484181*10^-14 +7.021621519*10^-17 -
4.38851344947*10^-16)]
161
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:
Result:
1.6181818182… result that is a very good approximation to the value of the golden
ratio 1.618033988749...
162
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
Planck’s Electric potential
1.042940*10
27
V Lorentz-Heaviside value
163
Relationship between Plancks 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
164
Fig. 1
It is therefore possible to consider the vortices of the "quantum vacuum" schematized
as cubes or octahedrons (the + sign inside a given vortex indicates its
counterclockwise rotation, while the - sign indicates its clockwise rotation). Between
vortex and vortex there is a layer of "bubbles"-universes (or universes-spheres),
which flows, as in the simplified two-dimensional drawing, from A to B
165
Fig. 2
Proposal
Image of space-time at quantum scale: the circles in red represent the points
corresponding to the compactified dimensions and the hexagons in blue, represent the
"fluctuations" (potential universes - green circles) of the quantum vacuum (2D). In
reality, we will have n-dimensional hyperspheres in which the compactified
dimensions "roll up" and octahedrons representing the "fluctuations", containing
spheres (bubbles of potential universes), of the quantum vacuum
166
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.
167
References
A NEW TYPE OF ISOTROPIC COSMOLOGICAL MODELS WITHOUT
SINGULARITY - A.A. STAROBINSKY - Volume 91B, number 1 PHYSICS
LETTERS 24 March 1980
A Number Theoretic Analysis of the Enthalpy, Enthalpy Energy Density,
Thermodynamic Volume, and the Equation of State of a Modified White Hole,
and the Implications to the Quantum Vacuum Spacetime, Matter Creation and
the Planck Frequency. - Nardelli, M., Kubeka, A.S. and Amani, A. (2024) - Journal
of Modern Physics , 15, 1-50. - https://doi.org/10.4236/jmp.2024.151001
Modular equations and approximations to - Srinivasa Ramanujan - Quarterly
Journal of Mathematics, XLV, 1914, 350 372
An Update on Brane Supersymmetry Breaking - Jihad Mourad and Augusto
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
168
See also:
The Geometry of the MRB constant by Marvin Ray Burns
https://www.academia.edu/22271085/The_Geometry_of_the_MRB_constant
(See also Page 29 the applications of the CMRB in various sectors of Theoretical
Physics (String Theory) and Cosmology )
http://xoom.virgilio.it/source_filemanager/na/ar/nardelli/michele%20and%20antonio
%20papers/Try%20to%20beat%20these%20MRB%20constant%20records!%20-
%20Online%20Technical%20Discussion%20Groups%E2%80%94Wolfram%20Com
munity%20b.pdf