1
On the analysis of some Ramanujan’s equations. New possible mathematical
connections with the Ramanujan Recurring Numbers, DN Constant and some
sectors of String Theory III
Michele Nardelli
1
, Antonio Nardelli
Abstract
In this paper (part III), we analyze some Ramanujan’s equations. We obtain new
possible mathematical connections with the Ramanujan Recurring Numbers, DN
Constant and some sectors of 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
3
From
Bruce Berndt – “Ramanujan: A Century of Inspiration”
https://www.youtube.com/watch?v=VRRRi1WcPJs&t=13s
We analyze the following equation.
We consider the right-hand side, for n = 2 and k = 4:
2^(4*2+2)*π^(4*2+3)*(-1)^5*((Bernoulli B(8))/((2*4)!))*((Bernoulli B(4))/(4*2+4-
2*4))-2*(4^(-4*2-3))/((e^(2π*4))-1)
Input
Exact result
4
Decimal approximation
-2.075513354313….
Alternate forms
Alternative representations
5
Series representations
6
Integral representations
From which, after some calculations:
-(2^(4*2+2)*π^(4*2+3)*(-1)^5*((Bernoulli B(8))/((2*4)!))*((Bernoulli
B(4))/(4*2+4-2*4))-2*(4^(-4*2-3))/((e^(2π*4))-1)+(sqrt((3 Φ)/10)))
Input
7
Exact result
Exact form
Decimal approximation
1.6449205111…. ≈ ζ(2) = π
2
/6 = 1.644934 (trace of the instanton shape and
Ramanujan Recurring Number)
Alternate forms
8
sqrt(6(-(2^(4*2+2)*π^(4*2+3)*(-1)^5*((Bernoulli B(8))/((2*4)!))*((Bernoulli
B(4))/(4*2+4-2*4))-2*(4^(-4*2-3))/((e^(2π*4))-1)+(sqrt((3 Φ)/10)))))
Input
Exact result
Exact form
Decimal approximation
3.14157970877…. ≈ π (Ramanujan Recurring Number)
Alternate forms
9
And again:
sqrt(1/(-(2^(4*2+2)*π^(4*2+3)*(-1)^5*((Bernoulli B(8))/((2*4)!))*((Bernoulli
B(4))/(4*2+4-2*4))-2*(4^(-4*2-3))/((e^(2π*4))-1)+(sqrt((3 Φ)/10))))*(4/3))
Input
Exact result
10
Exact form
Decimal approximation
0.90032…. ≈ 0.9003163161571…. =
(DN Constant)
Possible closed forms
Alternate forms
11
(6Pi*sqrt2)sqrt(1/(-(2^(4*2+2)*π^(4*2+3)*(-1)^5*((Bernoulli
B(8))/((2*4)!))*((Bernoulli B(4))/(4*2+4-2*4))-2*(4^(-4*2-3))/((e^(2π*4))-
1)+(sqrt((3 Φ)/10))))*(4/3))
Input
Exact result
Exact form
12
Decimal approximation
24.0000988914799…. ≈ 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). 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.
Alternate forms
13
((2Pi*sqrt2)sqrt(1/(-(2^(4*2+2)*π^(4*2+3)*(-1)^5*((Bernoulli
B(8))/((2*4)!))*((Bernoulli B(4))/(4*2+4-2*4))-2*(4^(-4*2-3))/((e^(2π*4))-
1)+(sqrt((3 Φ)/10))))*(4/3)))^4
Input
Exact result
Exact form
Decimal approximation
4096.06751033422925…. ≈ 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.
14
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)
Alternate forms
27*sqrt(((2Pi*sqrt2)sqrt(1/(-(2^(4*2+2)*π^(4*2+3)*(-1)^5*((Bernoulli
B(8))/((2*4)!))*((Bernoulli B(4))/(4*2+4-2*4))-2*(4^(-4*2-3))/((e^(2π*4))-
1)+(sqrt((3 Φ)/10))))*(4/3)))^4)+1
Input
15
Exact result
Exact form
Decimal approximation
1729.014240402449….
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, 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)
Alternate forms
16
Now, we have that:
for a = b = 0.5 and n = 4, we consider:
a^((n(n+1))/2)*b^((n(n-1))/2)
Input
Alternate form
Roots
17
Series expansion at n=0
Derivative
Indefinite integral
18
Series representations
From
We consider:
((q)^(n)^2)
Input
Values
19
3D plot (figure that can be related to a D-brane/Instanton)
The key observation from the above 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 appart.
Plots (figures that can be related to the open strings)
20
Contour plot
Root
Derivative
Indefinite integral
Series representations
21
Integral representation
From
we obtain:
q^(n(n+1)/2)
Input
22
Result
Values
3D plot (figure that can be related to a D-brane/Instanton)
The key observation from the above 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 appart.
23
Plots (figures that can be related to the open strings)
Contour plot
Alternate form
24
Roots
Derivative
Indefinite integral
Series representations
25
Integral representation
From
(-1)^n*q^((n(3n-1))/2)
Input
Values
26
Plots (figures that can be related to the open strings)
Alternate form
Alternate form assuming n and q are positive
Roots
27
Derivative
Indefinite integral
From the sum of the previous formulas, we obtain:
a^((n(n+1))/2)*b^((n(n-1))/2)+((q)^(n)^2)+q^(n(n+1)/2)+(-1)^n*q^((n(3n-1))/2)
Input
Alternate form
Alternate form assuming a, b, n, and q are positive
28
Derivative
Indefinite integral
From the indefinite integral result
we obtain:
q (a^(1/2 n (1 + n)) b^(1/2 (-1 + n) n) + q^(n^2)/(1 + n^2) + (2 q^(1/2 n (1 + n)))/(2 +
n + n^2) + (2 (-1)^n q^(1/2 n (-1 + 3 n)))/(2 - n + 3 n^2))
Input
Expanded forms
29
Alternate forms assuming a, b, n, and q are positive
Alternate forms
30
Derivative
Indefinite integral
Again, from the indefinite integral result
we obtain:
1/2 q^2 (a^(1/2 n (n+1)) b^(1/2 (n-1) n)+(8 q^(1/2 n (n+1)))/((n^2+n+2)
(n^2+n+4))+(8 (-1)^n q^(1/2 n (3n-1)))/((3n^2-n+2)(n(3n-1)+4))+(2
q^(n^2))/(n^4+3n^2+2))
31
Input
Expanded forms
Alternate forms assuming a, b, n, and q are positive
32
Alternate forms
33
Derivative
Indefinite integral
Multiplying the previous expression
By
for V = 1/3*√2*a^3 (octahedron volume) and V = (4/3*π*r^3) (sphere volume),
where r = (a/2), considering (1/3*√2*a^3) and (4/3*π*(a/2)^3) , we obtain:
34
(1/2 q^2 (a^(1/2 n (n+1)) b^(1/2 (n-1) n)+(8 q^(1/2 n (n+1)))/((n^2+n+2)
(n^2+n+4))+(8 (-1)^n q^(1/2 n (3n-1)))/((3n^2-n+2)(n(3n-1)+4))+(2
q^(n^2))/(n^4+3n^2+2)))(1/3*√2*a^3)
Input
Exact result
Expanded forms
35
Alternate forms assuming a, b, n, and q are positive
Alternate forms
36
Derivative
37
Indefinite integral
And:
(1/2 q^2 (a^(1/2 n (n+1)) b^(1/2 (n-1) n)+(8 q^(1/2 n (n+1)))/((n^2+n+2)
(n^2+n+4))+(8 (-1)^n q^(1/2 n (3n-1)))/((3n^2-n+2)(n(3n-1)+4))+(2
q^(n^2))/(n^4+3n^2+2)))(4/3*π*(a/2)^3)
Input
Exact result
38
Expanded forms
Alternate forms assuming a, b, n, and q are positive
39
Alternate forms
40
Series expansion at q=0
Series expansion at q=∞
Derivative
41
Indefinite integral
Dividing the two indefinite integral results
and
42
we obtain, simplifying:
1/(9√2)*1/(π/36)
Input
Result
Decimal approximation
0.9003163161571…. =
(DN Constant)
43
Property
Series representations
44
Now, we analyze the following equation.
We consider the left-hand side:
integrate(1/(sqrt(1-k^2*sin^2(ϕ))))dϕ, ϕ=0...π/2
Definite integral
Indefinite integral
45
From the definite integral result
we obtain:
EllipticK(k^2/(-1 + k^2))/sqrt(1 - k^2)
Input
Plots (figures that can be related to the open strings)
Alternate form
46
Series expansion at k=-1
Series expansion at k=0
Series expansion at k=1
Series expansion at k=∞
47
Derivative
From the Series expansion at k=∞
we obtain:
floor(arg(1/(1 - k^2))/(2 π)) (-(i k π)/(sqrt(-k^2) k) + ((1/k)^2)) + ((k (2 log(k) +
log(16) - i π))/(2 sqrt(-k^2) k) + ((1/k)^2))
Input
Exact result
48
Plots (figures that can be related to the open strings)
Alternate forms assuming k is real
Expanded form
49
Alternate forms assuming k>0
Alternate forms
Alternate form assuming k is positive
Numerical roots
50
Series expansion at k=0
51
Series expansion at k=∞
Derivative
Indefinite integral assuming all variables are real
From the indefinite integral result
we obtain:
52
-1/k - 1/2 (π + i log(16) + i log(k)) log(k) sgn(k)
Input
Plots (figures that can be related to the open strings)
Expanded form
Alternate form assuming k>0
53
Alternate forms
Numerical roots
Series expansion at k=0
Series expansion at k=∞
54
Derivative
Indefinite integral assuming all variables are real
And again, from the indefinite integral result
we obtain:
-log(k) + 1/2 k (π + i (-2 + log(16)) - (π + i (-2 + log(16))) log(k) - i log^2(k)) sgn(k)
Input
55
Plots (figures that can be related to the open strings)
Alternate form assuming k is real
Alternate forms
56
Expanded form
Alternate forms assuming k>0
Alternate form assuming k is positive
Numerical roots
57
Series expansion at k=0
Series expansion at k=∞
Derivative
Indefinite integral assuming all variables are real
58
From the indefinite integral result
for V = 1/3*√2*a^3 (octahedron volume) and V = (4/3*π*r^3) (sphere volume),
where r = (a/2), considering (1/3*√2*a^3) and (4/3*π*(a/2)^3) , we obtain:
(1/8 k (8 - 8 log(k) + k (-7 i + 3 π + 3 i log(16) - 2 (π + i (-3 + log(16))) log(k) - 2 i
log^2(k)) sgn(k)))(1/3*√2*a^3)
Input
Exact result
59
3D plots
Real part (figures that can be related to the D-branes/Instantons)
The key observation from the above 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 appart.
Imaginary part
60
Contour plots
Real part
61
Imaginary part
Alternate form assuming a and k are real
62
Alternate forms
Expanded form
Alternate forms assuming a and k are positive
63
Derivative
Indefinite integral
And:
(1/8 k (8 - 8 log(k) + k (-7 i + 3 π + 3 i log(16) - 2 (π + i (-3 + log(16))) log(k) - 2 i
log^2(k)) sgn(k)))(4/3*π*(a/2)^3)
Input
64
Exact result
3D plots
Real part (figures that can be related to the D-branes/Instantons)
The key observation from the above 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 appart.
65
Imaginary part
66
Contour plots
Real part
67
Imaginary part
Alternate form assuming a and k are real
Alternate forms
68
Expanded form
Alternate forms assuming a and k are positive
69
Derivative
Indefinite integral
Alternative representations
70
Dividing the two indefinite integral results
and
we obtain, simplifying:
71
1/(48√2)*1/(π/192)
Input
Result
Decimal approximation
0.9003163161571…. =
(DN Constant)
Property
Series representations
72
Now, we have:
From
We consider:
(((k^2)^3)/(ℓ^2))^(1/8)-(((1-k^2)^3)/(1-ℓ^2))^(1/8) – 1
Input
73
3D plots
Real part (figures that can be related to the D-branes/Instantons)
The key observation from the above 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 appart.
74
Imaginary part
Contour plots
Real part
75
Imaginary part
Alternate forms assuming k and ℓ are positive
Alternate form
76
Alternate form assuming k and ℓ are positive
Property as a function
Parity
Series expansion at k=0
Series expansion at k=∞
Derivative
77
Indefinite integral
and:
1 - (((1-ℓ^2)^3)/(1-k^2))^(1/8)+(((ℓ^2)^3)/(k^2))^(1/8)
Input
3D plot
78
Contour plot
The key observation from the above 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 appart.
Alternate forms assuming k and ℓ are positive
79
Alternate form
Property as a function
Parity
Series expansion at k=0
Series expansion at k=∞
Derivative
80
Indefinite integral
From the algebraic sum of the two previous expressions
and
we obtain:
((k^6/ℓ^2)^(1/8) - ((k^2 - 1)^3/(ℓ^2 - 1))^(1/8) - 1)-((ℓ^6/k^2)^(1/8) - ((ℓ^2 -
1)^3/(k^2 - 1))^(1/8) + 1)
Input
81
Result
Alternate form assuming k and ℓ are positive
Property as a function
Parity
Series expansion at k=0
Series expansion at k=∞
82
Derivative
Indefinite integral
From the result
for V = 1/3*√2*a^3 (octahedron volume) and V = (4/3*π*r^3) (sphere volume),
where r = (a/2), considering (1/3*√2*a^3) and (4/3*π*(a/2)^3) , we obtain:
((k^6/ℓ^2)^(1/8) - (ℓ^6/k^2)^(1/8) - ((k^2 - 1)^3/(ℓ^2 - 1))^(1/8) + ((ℓ^2 - 1)^3/(k^2 -
1))^(1/8) - 2)(1/3*√2*a^3)
83
Input
Exact result
Alternate form
Expanded form
Property as a function
Parity
84
Derivative
Indefinite integral
And:
((k^6/ℓ^2)^(1/8) - (ℓ^6/k^2)^(1/8) - ((k^2 - 1)^3/(ℓ^2 - 1))^(1/8) + ((ℓ^2 - 1)^3/(k^2 -
1))^(1/8) - 2)(4/3*π*(a/2)^3)
Input
Exact result
85
Alternate form
Expanded form
Property as a function
Parity
Derivative
Indefinite integral
86
Dividing the two indefinite integral results
and
we obtain, simplifying:
-1/(3√2)*1/(-π/12)
Input
Result
87
Decimal approximation
0.9003163161571…. =
(DN Constant)
Property
Series representations
88
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
89
Result
Decimal approximation
0.900316316157106…. =
(DN Constant)
Property
Series representations
90
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)
91
Property
Series representations
Integral representations
92
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
93
Decimal approximation
1.64529737852…. ≈ ζ(2) = π
2
/6 = 1.644934 (trace of the instanton shape and
Ramanujan Recurring Number)
Alternate forms
Expanded forms
94
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
95
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
96
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
97
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
98
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.
99
Indeed:
From:
https://www.academia.edu/75787512/The_Theory_of_String_A_Candidate_for_a_G
eneralized_Unification_Model
The Einstein’s field equation and the String Theory.
The Einstein’s field equation which includes the cosmological constant is:
μν
μν
μν
GT
μν
(8)
where
μν
is the Ricci tensor, its trace, the cosmological constant,
μν
the metric
tensor of the space geometry, G the Newton’s gravitational constant and
μν
the tensor
representing the properties of energy, matter and momentum.
The left hand-side of (8) represents the gravitational field and, consequently, the
warped space-time, while the right hand-side represents the matter, i.e. the sources of
the gravitational field.
In string theory the gravity is related to the gravitons which are bosons, whereas the
matter is related to fermions. It follows that the left and right hand of (8) may be
respectively related to the action of bosonic and of superstrings.
The actions of bosonic string and superstring (also containing fermions) are
connected by the Palumbo-Nardelli relation (Palumbo et al. 2005):
100
The sign minus in the above equation comes from the inversion of any relationship.
(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
101
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, 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
102
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
103
Series representations
104
From
we obtain also:
sqrt(6(1/3*(2/(((2sqrt2)/Pi)))^2))
Input
105
Exact result
Decimal approximation
3.14159265358… = π
Property
All 2
nd
roots of π
2
Series representations
106
Integral representations
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
107
From
, we obtain:
sqrt(1/(Pi^2/6)*(4/3))
Input
Exact result
Decimal approximation
0.900316316157106…. =
(DN Constant)
Property
All 2
nd
roots of 8/π
2
108
Series representations
109
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
110
“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
111
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 Approximations 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
112
We have, in certain cases, the following connections:
Fig. 1
Fig. 2
113
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
115
Where ζ(2+it) :
Input
Plots
Roots
116
Series expansion at t=0
Alternative representations
Series representations
117
Integral representations
Functional equations
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:
118
we obtain:
2Pi/(ln(2))
Input:
Exact result:
Decimal approximation:
9.06472028365….
Alternative representations:
119
Series representations:
Integral representations:
120
From which:
(2Pi/(ln(2)))*(1/12 π log(2))
Input:
Exact result:
Decimal approximation:
1.6449340668…. = ζ(2) =
121
From:
Modular equations and approximations to - Srinivasa Ramanujan - Quarterly
Journal of Mathematics, XLV, 1914, 350 – 372
We have that:
122
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
123
Minimal polynomial
Expanded forms
Series representations
124
125
Or:
27((4096+276)^1/2-2-1/2(((√(10-2√5) -2))⁄((√5-1))))+φ
Input
Result
Decimal approximation
1729.0526944…. as above
Alternate forms
126
Minimal polynomial
Expanded forms
Series representations
127
128
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
129
Minimal polynomial
Expanded forms
All 15th roots of ϕ + 27 (-2 + 2 sqrt(1093) - (sqrt(10 - 2 sqrt(5)) - 2)/(2 (sqrt(5) -
1)))
130
Series representations
131
Integral representation
132
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:
133
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:
134
Now, we have the following calculations:
= 1.6272016… * 10^-6
from which:
= 1.6272016… * 10^-6
0.000244140625
=
= 1.6272016… * 10^-6
Now:
And:
135
(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:
136
Result:
0.00666501785…
Series representations:
Now:
= 0.0066650177536
=
= 0.00666501785…
137
From:
ln(0.00666501784619)
Input interpretation:
Result:
-5.010882647757…
Alternative representations:
Series representations:
138
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:
139
(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:
140
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:
141
(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:
142
Solutions:
Alternate forms:
Expanded form:
Alternate form assuming b, M, and V are positive:
Alternate form assuming b, M, and V are real:
143
Derivative:
Implicit derivatives
Global minimum:
144
Global minima:
From:
we obtain:
(225.913 (-0.054323 M^2 + 6.58545×10^-10 sqrt(M^4)))/M^2
Input interpretation:
145
Result:
Plots:
Alternate form assuming M is real:
-12.2723 result very near to the black hole entropy value 12.1904 = ln(196884)
Alternate forms:
146
Expanded form:
Property as a function:
Parity
Series expansion at M = 0:
Series expansion at M = ∞:
Derivative:
147
Indefinite integral:
Global maximum:
Global minimum:
Limit:
Definite integral after subtraction of diverging parts:
148
From b that is equal to
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:
149
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
Root:
Property as a function:
Parity
150
Series expansion at M = 0:
Series expansion at M = ∞:
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:
151
Result:
-4.38851344947*10
-16
For M = 0.2:
1/3 (0.0814845 ((225.913 (-0.054323 0.2^2 + 6.58545×10^-10 sqrt(0.2^4)))/0.2^2 ) +
1)^2 0.2^2
Input interpretation:
Result:
7.021621519159*10
-17
152
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
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
153
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)]
Input interpretation:
154
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...
155
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
156
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
157
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
158
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
159
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
160
References
Bruce Berndt – “Ramanujan: A Century of Inspiration”
https://www.youtube.com/watch?v=VRRRi1WcPJs&t=13s
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 - Emilian Dudas, Sergio Ferrara, Alex
Kehagias and Augusto Sagnotti - arXiv:1507.07842v2 [hep-th] 14 Sep 2015
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