on the study of five ramanujan mock theta functions. new possible mathematical connections with some...
DESCRIPTION
In this paper, we analyze five Ramanujan Mock Theta functions. We describe the new possible mathematical connections with some cosmological and physical parametersTRANSCRIPT
1
On the study of five Ramanujan Mock Theta functions. New possible
mathematical connections with some cosmological and physical parameters.
Michele Nardelli1, Antonio Nardelli
2
Abstract
In this paper, we analyze five Ramanujan Mock Theta functions. We describe the new
possible mathematical connections with some cosmological and physical 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
2
From:
THE MOCK THETA FUNCTIONS (2) - By G. N. WATSON. [Received 3
August, 1936.—Read 12 November, 1936]
Now, we have the following two mock theta functions:
For q = 2, we obtain:
q+q^3(1+q)+q^6(1+q)(1+q^2)+q^10(1+q)(1+q^2)(1+q^3)
Input
Plots (figures that can be related to the open strings)
7
From the solution of the integral
we obtain, for q = 2 :
q^17/17 + q^16/16 + q^15/15 + q^14/7 + q^13/13 + q^12/12 + q^11/11 + q^10/10 +
q^9/9 + q^8/8 + q^7/7 + q^5/5 + q^4/4 + q^2/2
2^17/17 + 2^16/16 + 2^15/15 + 2^14/7 + 2^13/13 + 2^12/12 + 2^11/11 + 2^10/10 +
2^9/9 + 2^8/8 + 2^7/7 + 2^5/5 + 2^4/4 + 2^2/2
Input
Exact result
Decimal approximation
17710.8660098….
From:
we obtain:
8
1+((q^2)/(1-q))+((q^8)/((1-q)(1-q^3)))
Input
Plots (figures that can be related to the open strings)
Alternate forms
11
From the solution of the integral
we obtain, for q = 2:
q^5/5 + q^4/4 + q^3/3 + q^2/2 - 1/6 log(q^2 + q + 1) + 2 q + 1/(3 - 3 q) + 4/3 log(1 -
q) + (tan^(-1)((2 q + 1)/sqrt(3)))/(3 sqrt(3)) + 3/2
2^5/5 + 2^4/4 + 2^3/3 + 2^2/2 - 1/6 log(2^2 + 2 + 1) + 2*2 + 1/(3 – 3*2) + 4/3 log(1
- 2) + (tan^(-1)((2*2 + 1)/sqrt(3)))/(3 sqrt(3)) + 3/2
Input
18
Dividing the two exact results of the above integrals, we obtain:
((2712472262/153153))*1/((607/30 + (4 i π)/3 + (tan^(-1)(5/sqrt(3)))/(3 sqrt(3)) -
log(7)/6))
Input
Exact Result
Decimal approximation
26
Multiplying the two exact solutions, we obtain:
((2712472262/153153))*((607/30 + (4 i π)/3 + (tan^(-1)(5/sqrt(3)))/(3 sqrt(3)) -
log(7)/6))
Input
34
And from the difference and sum, we obtain:
((2712472262/153153))-((607/30 + (4 i π)/3 + (tan^(-1)(5/sqrt(3)))/(3 sqrt(3)) -
log(7)/6))
Input
40
((2712472262/153153))+((607/30 + (4 i π)/3 + (tan^(-1)(5/sqrt(3)))/(3 sqrt(3)) -
log(7)/6))
Input
47
From which:
1/2((((2712472262/153153))*((607/30+(4iπ)/3+(tan^(-1)(5/sqrt(3)))/(3sqrt(3)) -
log(7)/6))))+((27155710577/1531530+(4iπ)/3+(tan^(-1)(5/sqrt(3)))/(3sqrt(3)) -
log(7)/6))-(2207+322+123+29+7)-(76+18+4)
Input
Exact Result
Decimal approximation
48
Polar coordinates
196883
196884/196883 is a fundamental number of the following j-invariant
(In mathematics, Felix Klein's j-invariant or j function, regarded as a function of
a complex variable τ, is a modular function of weight zero for SL(2, Z) defined on
the upper half plane of complex numbers. Several remarkable properties of j have to
do with its q expansion (Fourier series expansion), written as a Laurent series in
terms of q = e2πiτ
(the square of the nome), which begins:
Note that j has a simple pole at the cusp, so its q-expansion has no terms below q−1
.
All the Fourier coefficients are integers, which results in several almost integers,
notably Ramanujan's constant:
The asymptotic formula for the coefficient of qn is given by
as can be proved by the Hardy–Littlewood circle method)
Furthermore, 196884 is the coefficient of q of the partition function Z1(q) that is the
number of quantum states of the minimal black hole for the value of k equal to 1.
58
Furthermore, from the ratio of the two exact results of the previous integrals
860.67
we obtain also:
59
2(((2712472262/153153))*1/((607/30 + (4 i π)/3 + (tan^(-1)(5/sqrt(3)))/(3 sqrt(3)) -
log(7)/6)))+8
Input
Exact Result
Decimal approximation
Polar coordinates
1729.2
This result is very near to the mass of candidate glueball f0(1710) scalar meson.
Furthermore, 1728 occurs in the algebraic formula for the j-invariant of an elliptic
curve. (1728 = 82
* 33) The number 1728 is one less than the Hardy–Ramanujan
number 1729 (taxicab number)
Polar forms
67
(1/27(2(((2712472262/153153))*1/((607/30 + (4 i π)/3 + (tan^(-1)(5/sqrt(3)))/(3
sqrt(3)) - log(7)/6)))+7))^2-1
Input
68
Exact Result
Decimal approximation
Polar coordinates
4096 = 642 where 4096 and 64 are fundamental values indicated in the Ramanujan
paper ―Modular equations and Approximations to π‖
77
(2(((2712472262/153153))*1/((607/30 + (4 i π)/3 + (tan^(-1)(5/sqrt(3)))/(3 sqrt(3)) -
log(7)/6)))+8)^1/15
Input
Exact Result
Decimal approximation
Polar coordinates
1.6438 ≈ ζ(2) = 𝜋2
6= 1.644934… (trace of the instanton shape)
80
All 15th roots of 8 + 5424944524/(153153 (607/30 + (4 i π)/3 - log(7)/6 + (tan^(-
1)(5/sqrt(3)))/(3 sqrt(3))))
91
From the initial expression, we calculate the following integrals:
integrate(2^5/5 + 2^4/4 + 2^3/3 + 2^2/2 - 1/6 log(2^2 + 2 + 1) + 2*2 + 1/(3 – 3*2) +
4/3 log(1 - 2) + (tan^(-1)((2*2 + 1)/sqrt(3)))/(3 sqrt(3)) + 3/2)x
Indefinite integral
Plot of the integral (figure that can be related to an open string)
Alternate forms of the integral
92
From the result:
integrate(2/3 i π x^2 + (607 x^2)/60 - 1/12 x^2 log(7) + (x^2 tan^(-1)(5/sqrt(3)))/(6
sqrt(3)))x
Indefinite integral
Plot of the integral (figure that can be related to an open string)
93
Alternate forms of the integral
Expanded form of the integral
From the result:
integrate(1/720 x^4 (1821 + 120 i π - 15 log(7) + 10 sqrt(3) tan^(-1)(5/sqrt(3))))x
Indefinite integral
94
Plot of the integral (figure that can be related to an open string)
Alternate forms of the integral
From the result, for x = 1, we obtain:
(1/36 i π + (607)/1440 - 1/288 log(7) + (tan^(-1)(5/sqrt(3)))/(144 sqrt(3)))
Input
99
From the exact result
we obtain:
607/1440 + (i π)/36 + (tan^(-1)(5/sqrt(3)))/(x* sqrt(3)) - log(7)/288 =
0.419732040151544 + 0.08726646259971647i
Input interpretation
100
Result
Alternate form assuming x is real
Alternate forms
Real solution
Complex solution
144 (Fibonacci number)
101
Thence, we obtain:
1/((((sqrt(3))*((0.41973204015 + 0.08726646259i-(607/1440 + (i π)/36)+
(log(7)/288))) *1/((tan^(-1)(5/sqrt(3)))))))
Input interpretation
Result
Polar coordinates
144
110
From which, we obtain:
12*1/((((sqrt(3))*((0.41973204015 + 0.08726646259i-(607/1440 + (i π)/36)+
(log(7)/288))) *1/((tan^(-1)(5/sqrt(3)))))))+1
Input interpretation
Result
Polar coordinates
1729
This result is very near to the mass of candidate glueball f0(1710) scalar meson.
Furthermore, 1728 occurs in the algebraic formula for the j-invariant of an elliptic
curve. (1728 = 82
* 33) The number 1728 is one less than the Hardy–Ramanujan
number 1729 (taxicab number)
Polar forms
119
(12*1/((((sqrt(3))*((0.41973204015 + 0.08726646259i-(607/1440 + (i π)/36)+
(log(7)/288))) *1/((tan^(-1)(5/sqrt(3)))))))+1)^1/15
Input interpretation
Result
Polar coordinates
1.64382 ≈ ζ(2) = 𝜋2
6= 1.644934… (trace of the instanton shape)
120
From:
S. Ramanujan to G.H. Hardy - 12 January 1920 - University of Madras
Now we have the following three mock theta functions:
We analyze these functions and consider the following data:
; t = 0.25 ; q = 0.7788
From:
we obtain:
1+(q/(1+q^2))+((q^4)/((1+q)(1+q^2)))+((q^9)/((1+q)(1+q^2)(1+q^3)))
Input
124
we obtain:
derivative((3 q^13 + q^12 + 5 q^11 + 9 q^10 + 12 q^8 + q^7 + 5 q^6 + q^5 + q^4 + 5
q^3 - q^2 + (1 + q))/((1 + q)^3 (1 + q^2)^2 ((1 + q^2) - q)^2))
Derivative
Plots (figures that can be related to the open strings)
Alternate forms
127
Local maxima
Local minimum
From the result of the above alternate form :
For q = 0.7788 , we obtain:
(2*0.7788+3)/(0.7788^2+1)^2-(4(2*0.7788+1))/(0.7788^2+1)^3-
(2*0.7788)/((0.7788-1) 0.7788+1)^3+6*0.7788+(2(0.7788+1))/(3((0.7788-1)
0.7788+1)^2)+10/(3 (0.7788+1)^3)-1/(0.7788+1)^4 – 2
128
Input
Result
1.4479789895….
From which:
((2*0.7788+3)/(0.7788^2+1)^2-(4(2*0.7788+1))/(0.7788^2+1)^3-
(2*0.7788)/((0.7788-1) 0.7788+1)^3+6*0.7788+(2(0.7788+1))/(3((0.7788-1)
0.7788+1)^2)+10/(3 (0.7788+1)^3)-1/(0.7788+1)^4 - 2)^23-(843+47-3)
Input
Result
4096.47156853429….. ≈ 4096 = 642, where 4096 and 64 are fundamental values
indicated in the Ramanujan paper ―Modular equations and Approximations to π‖
129
27√(((2*0.7788+3)/(0.7788^2+1)^2-(4(2*0.7788+1))/(0.7788^2+1)^3-
(2*0.7788)/((0.7788-1)0.7788+1)^3+6*0.7788+(2(0.7788+1))/(3((0.7788-
1)0.7788+1)^2)+10/(3(0.7788+1)^3)-1/(0.7788+1)^4-2)^23-(843+47-3))+1
Input
Result
1729.10….
This result is very near to the mass of candidate glueball f0(1710) scalar meson.
Furthermore, 1728 occurs in the algebraic formula for the j-invariant of an elliptic
curve. (1728 = 82
* 33) The number 1728 is one less than the Hardy–Ramanujan
number 1729 (taxicab number)
(27√(((2*0.7788+3)/(0.7788^2+1)^2-(4(2*0.7788+1))/(0.7788^2+1)^3-
(2*0.7788)/((0.7788-1)0.7788+1)^3+6*0.7788+(2(1.7788))/(3((0.7788-
1)0.7788+1)^2)+10/(3(1.7788)^3)-1/(1.7788)^4-2)^23-(843+44))+1)^1/15
Input interpretation
130
Result
1.643821533….≈ ζ(2) = 𝜋2
6= 1.644934… (trace of the instanton shape)
Now, from the solution of the above integral:
we obtain:
1/36 (3 (3 q^4 - 4 q^3 + 6 q^2 + 6 log(q^2 + 1) + 2 log(q^2 - q + 1) - 12 q + 2/(q + 1)
+ 20 log(q + 1)) + 18 tan^(-1)(q) + 4 sqrt(3) tan^(-1)((2 q - 1)/sqrt(3)))
for q = 0.7788 :
1/36 (3 (3 0.7788^4 - 4 0.7788^3 + 6 0.7788^2 + 6 log(0.7788^2 + 1) + 2
log(0.7788^2 – 0.7788 + 1) - 12 0.7788 + 2/(0.7788 + 1) + 20 log(0.7788 + 1)) + 18
tan^(-1)(0.7788) + 4 sqrt(3) tan^(-1)((2 0.7788 - 1)/sqrt(3)))
132
1/((1/36(3(3 0.7788^4-4 0.7788^3+6 0.7788^2+6log(0.7788^2+1)+2log(0.7788^2–
0.7788+1)-12 0.7788+2/(0.7788+1)+20log(0.7788+1))+18tan^(-
1)(0.7788)+4sqrt(3)tan^(-1)((2 0.7788-1)/sqrt(3))))-0.5)
133
Input
Result
1.6423616598…..≈ ζ(2) = 𝜋2
6= 1.644934… (trace of the instanton shape)
Alternative representations
141
Now, we analyze the second mock theta function:
1+q(1+q)+q^4(1+q)(1+q^3)+q^9(1+q)(1+q^3)(1+q^5)
Input
Plots (figures that can be related to the open strings)
143
Derivative
Indefinite integral
Local minimum
From:
Perform the derivative, we obtain:
derivative( 18 q^17 + 17 q^16 + 15 q^14 + 14 q^13 + 13 q^12 + 12 q^11 + 10 q^9 +
9 q^8 + 8 q^7 + 7 q^6 + 5 q^4 + 4 q^3 + 2 q + 1)
146
Local maximum
Local minima
From:
2 (153 q^16 + 136 q^15 + 105 q^13 + 91 q^12 + 78 q^11 + 66 q^10 + 45 q^8 + 36
q^7 + 28 q^6 + 21 q^5 + 10 q^3 + 6 q^2 + 1)
For q = 0.7788 , we obtain:
2 (153 0.7788^16 + 136 0.7788^15 + 105 0.7788^13 + 91 0.7788^12 + 78 0.7788^11
+ 66 0.7788^10 + 45 0.7788^8 + 36 0.7788^7 + 28 0.7788^6 + 21 0.7788^5 + 10
0.7788^3 + 6 0.7788^2 + 1)
147
Input
Result
117.9583107512….
From which:
16(2 (153 0.7788^16 + 136 0.7788^15 + 105 0.7788^13 + 91 0.7788^12 + 78
0.7788^11 + 66 0.7788^10 + 45 0.7788^8 + 36 0.7788^7 + 28 0.7788^6 + 21
0.7788^5 + 10 0.7788^3 + 6 0.7788^2 + 1)-Pi^2)-√2
Input
Result
1728.01….
This result is very near to the mass of candidate glueball f0(1710) scalar meson.
Furthermore, 1728 occurs in the algebraic formula for the j-invariant of an elliptic
curve. (1728 = 82
* 33) The number 1728 is one less than the Hardy–Ramanujan
number 1729 (taxicab number)
148
Series representations
(1/27(16(2 (153 0.7788^16 + 136 0.7788^15 + 105 0.7788^13 + 91 0.7788^12 + 78
0.7788^11 + 66 0.7788^10 + 45 0.7788^8+36 0.7788^7+28 0.7788^6+21
0.7788^5+10 0.7788^3+6 0.7788^2+1)-Pi^2)-√2))^2
149
Input
Result
4096.02….≈ 4096 = 642 where 4096 and 64 are fundamental values indicated in the
Ramanujan paper ―Modular equations and Approximations to π‖
((16(2 (153 0.7788^16+136 0.7788^15+105 0.7788^13+91 0.7788^12+78
0.7788^11+66 0.7788^10 + 45 0.7788^8 + 36 0.7788^7 + 28 0.7788^6 + 21 0.7788^5
+ 10 0.7788^3 + 6 0.7788^2 + 1)-Pi^2)-√2)+1)^1/15
Input
150
Result
1.64382….≈ ζ(2) = 𝜋2
6= 1.644934… (trace of the instanton shape)
Now, from the solution of the above integral:
q^19/19 + q^18/18 + q^16/16 + q^15/15 + q^14/14 + q^13/13 + q^11/11 + q^10/10 +
q^9/9 + q^8/8 + q^6/6 + q^5/5 + q^3/3 + q^2/2 + q
For q = 0.7788 , we obtain:
(0.7788^19/19+0.7788^18/18+0.7788^16/16+0.7788^15/15+0.7788^14/14+0.7788^1
3/13+0.7788^11/11+0.7788^10/10+0.7788^9/9+0.7788^8/8+0.7788^6/6+0.7788^5/5
+0.7788^3/3+0.7788^2/2+0.7788)
Input
Result
1.3855807917….
151
1+1/(3(1/e(0.7788^19/19+0.7788^18/18+0.7788^16/16+0.7788^15/15+0.7788^14/14
+0.7788^13/13+0.7788^11/11+0.7788^10/10+0.7788^9/9+0.7788^8/8+0.7788^6/6+
0.7788^5/5+0.7788^3/3+0.7788^2/2+0.7788)))
Input
Result
1.65394522514…. result very near to the 14th root of the following Ramanujan’s
class invariant 𝑄 = 𝐺505/𝐺101/5 3 = 1164.2696 i.e. 1.65578...
Alternative representation
153
we obtain:
((1 + (q (1 + q + 2 q^3 + q^4 + q^6 + q^8))/(1 + q + q^2 + 2 q^3 + q^4 + q^5 +
q^6)))+((1 + q (1 + q) (1 + q^3 + q^6 + q^8 + q^11 + q^13 + q^16)))
Input
Result
Plots (figures that can be related to the open strings)
155
Derivative
From the expression
For q = 0.7788 , we obtain:
q (q + 1) (q^16 + q^13 + q^11 + q^8 + q^6 + q^3 + 1) + (q (q^8 + q^6 + q^4 + 2 q^3
+ q + 1))/(q^6 + q^5 + q^4 + 2 q^3 + q^2 + q + 1) + 2
0.7788 (0.7788+1)
(0.7788^16+0.7788^13+0.7788^11+0.7788^8+0.7788^6+0.7788^3+1)+(0.7788(0.77
88^8+0.7788^6+0.7788^4+2
0.7788^3+0.7788+1))/(0.7788^6+0.7788^5+0.7788^4+2
0.7788^3+0.7788^2+0.7788+1)+2
Input
156
Result
5.3425012228441….
11((0.7788(1.7788)(0.7788^16+0.7788^13+0.7788^11+0.7788^8+0.7788^6+1.47236
)+(0.7788(0.7788^8+0.7788^6+0.7788^4+2
0.47236+1.7788)/(0.7788^6+0.7788^5+0.7788^4+2
0.47236+0.7788^2+1.7788)+2))^3)+47+2+φ^2
Input interpretation
Result
1728.97….
This result is very near to the mass of candidate glueball f0(1710) scalar meson.
Furthermore, 1728 occurs in the algebraic formula for the j-invariant of an elliptic
curve. (1728 = 82
* 33) The number 1728 is one less than the Hardy–Ramanujan
number 1729 (taxicab number)
158
(11((0.7788(1.7788)(0.7788^16+0.7788^13+0.7788^11+0.7788^8+0.7788^6+1.4723
6)+(0.7788(0.7788^8+0.7788^6+0.7788^4+2.72352)/(0.7788^6+0.7788^5+0.7788^4
+0.94472+0.7788^2+1.7788)+2))^3)+7^2+φ^2)^1/15
Input interpretation
Result
1.64381346388….≈ ζ(2) = 𝜋2
6= 1.644934… (trace of the instanton shape)
159
Now, we analyze the third mock theta function:
We obtain:
1+[q/(1-q)+(q^3)/((1-q^2)(1-q^3))+(q^5)/((1-q^3)(1-q^4)(1-q^5))]
Input
Result
Plots (figures that can be related to the open strings)
162
Limit
From
for q = 0.7788 :
1 + 0.7788/(1 - 0.7788) + 0.7788^3/((1 - 0.7788^2) (1 - 0.7788^3)) + 0.7788^5/((1 -
0.7788^3) (1 - 0.7788^4) (1 - 0.7788^5))
Input
Result
7.999997103998…. ≈ 8
From which:
(1 + 0.7788/(1 - 0.7788) + 0.7788^3/((1 - 0.7788^2) (1 - 0.7788^3)) + 0.7788^5/((1 -
0.7788^3) (1 - 0.7788^4) (1 - 0.7788^5)))^4
163
Input
Result
4095.99406899….≈ 4096 = 642 where 4096 and 64 are fundamental values indicated
in the Ramanujan paper ―Modular equations and Approximations to π‖
27sqrt((1 + 0.7788/(1 - 0.7788) + 0.7788^3/((1 - 0.7788^2) (1 - 0.7788^3)) +
0.7788^5/((1 - 0.7788^3) (1 - 0.7788^4) (1 - 0.7788^5)))^4)+1
Input
164
Result
1728.9987489…..≈ 1729
This result is very near to the mass of candidate glueball f0(1710) scalar meson.
Furthermore, 1728 occurs in the algebraic formula for the j-invariant of an elliptic
curve. (1728 = 82
* 33) The number 1728 is one less than the Hardy–Ramanujan
number 1729 (taxicab number)
(27sqrt((1 + 0.7788/(1 - 0.7788) + 0.7788^3/((1 - 0.7788^2) (1 - 0.7788^3)) +
0.7788^5/((1 - 0.7788^3) (1 - 0.7788^4) (1 - 0.7788^5)))^4)+1)^1/15
Input
Result
1.643815149453…..≈ ζ(2) = 𝜋2
6= 1.644934… (trace of the instanton shape)
We have also:
((1 + 0.7788/(1 - 0.7788) + 0.7788^3/((1 - 0.7788^2) (1 - 0.7788^3)) + 0.7788^5/((1 -
0.7788^3) (1 - 0.7788^4) (1 - 0.7788^5))))^136*(tan^2(17/(5^2*3)))
where
165
Input
Result
0.351600…*10
122 ≈ ΛQ
The observed value of ρΛ or Λ today is precisely the classical dual of its quantum
precursor values ρQ , ΛQ in the quantum very early precursor vacuum UQ as
determined by our dual equations
Alternative representations
168
Multiple-argument formulas
From the derivative of
Performing:
second derivative of (1 + (q/(1 - q) + q^3/((1 - q^2) (1 - q^3)) + q^5/((1 - q^3) (1 -
q^4) (1 - q^5))))
172
From the alternate form
For q = 0.7788 , we obtain:
-(2 (0.7788^29 + 4*0.7788^28 + 16*0.7788^27 + 41*0.7788^26 + 80*0.7788^25 +
113*0.778^24 + 133*0.7788^23 + 88*0.7788^22 – 44*0.7788^21 – 307*0.7788^20 -
655 q^19 - 988 q^18 - 1078 q^17 - 781 q^16 + 14 q^15 + 1149 q^14 + 2379 q^13 +
3327 q^12 + 3841 q^11 + 3817 q^10 + 3385 q^9 + 2691 q^8 + 1932 q^7 + 1233 q^6
+ 695 q^5 + 332 q^4 + 131 q^3 + 40 q^2 + 10 q + 1))
Input
Result
-(2 (-1.004035093647- 655 0.7788^19 - 988 0.7788^18 - 1078 0.7788^17 - 781
0.7788^16 + 14 0.7788^15 + 1149 0.7788^14 + 2379 0.7788^13 + 3327 0.7788^12 +
3841 0.7788^11 + 3817 0.7788^10 + 3385 q^9 + 2691 q^8 + 1932 q^7 + 1233 q^6 +
695 q^5 + 332 q^4 + 131 q^3 + 40 q^2 + 10 q + 1))
173
Input
Result
-(2 (-1.004035093647- 816.77764665+3385 0.7788^9 + 2691 0.7788^8 + 1932
0.7788^7 + 1233 0.7788^6 + 695 0.7788^5 + 332 0.7788^4 + 131 0.7788^3 + 40
0.7788^2 + 10 0.7788 + 1))
/((0.7788 - 1)^5 (0.7788 + 1)^3 (0.7788^2 + 1)^3 (0.7788^2 + 0.7788 + 1)^3
(0.7788^4 + 0.7788^3 + 0.7788^2 + 0.7788 + 1)^3)
Input
Result
Thence, in conclusion:
-(2 (-1.004035093647- 816.77764665+3385 0.7788^9 + 2691 0.7788^8 + 1932
0.7788^7 + 1233 0.7788^6 + 695 0.7788^5 + 332 0.7788^4 + 131 0.7788^3 + 40
0.7788^2 + 10 0.7788 + 1))/(-5.629095041)
174
Input interpretation
Result
330.4996005811….
From which:
2e(-(2 (-1.004035093647- 816.77764665+3385 0.7788^9 + 2691 0.7788^8 + 1932
0.7788^7 + 1233 0.7788^6 + 695 0.7788^5 + 332 0.7788^4 + 131 0.7788^3 + 40
0.7788^2 + 10 0.7788 + 1))/(-5.629095041))-64-2-e
Input interpretation
Result
1728.06….
This result is very near to the mass of candidate glueball f0(1710) scalar meson.
Furthermore, 1728 occurs in the algebraic formula for the j-invariant of an elliptic
curve. (1728 = 82
* 33) The number 1728 is one less than the Hardy–Ramanujan
number 1729 (taxicab number)
176
Performing the 15th root of 1729.063853…, we obtain:
(2e(-(2 (-1.004035093647- 816.77764665+3385 0.7788^9 + 2691 0.7788^8 + 1932
0.7788^7+1233 0.7788^6+695 0.7788^5+332 0.7788^4+131 0.7788^3+40
0.7788^2+10 0.7788+1))/(-5.629095041))-64-2-e+1)^1/15
Input interpretation
Result
1.6438192746….≈ ζ(2) = 𝜋2
6= 1.644934… (trace of the instanton shape)
(1/27(2e(-(2 (-1.004035093647- 816.77764665+3385 0.7788^9 + 2691 0.7788^8 +
1932 0.7788^7+1233 0.7788^6+695 0.7788^5+332 0.7788^4+131 0.7788^3+40
0.7788^2+10 0.7788+1))/(-5.629095041))-64-2-e))^2
177
Input interpretation
Result
4096.3….≈ 4096 = 642 where 4096 and 64 are fundamental values indicated in the
Ramanujan paper ―Modular equations and Approximations to π‖
From the above integral
178
(-144 sqrt(5) log(-2 q^2 + (sqrt(5) - 1) q - 2) + 450 log(q^2 + 1) - 400 log(q^2 + q +
1) + 144 sqrt(5) log(2 q^2 + sqrt(5) q + q + 2) - 570/(q - 1) + 30/(q - 1)^2 - 3475
log(1 - q) + 900 log(q - 1) - 1125 log(q + 1) + 800 sqrt(3) tan^(-1)((2 q +
1)/sqrt(3)))/3600
-144 sqrt(5) log(-2 0.7788^2 + (sqrt(5) - 1) 0.7788 - 2) + 450 log(0.7788^2 + 1) - 400
log(0.7788^2 + 0.7788 + 1) + 144 sqrt(5) log(2 0.7788^2 + sqrt(5) 0.7788 + 0.7788 +
2)
Input
Result
Polar coordinates
1025.22
(((166.722 -1011.57 i) - 570/(0.7788 - 1) + 30/(0.7788 - 1)^2 - 3475 log(1 – 0.7788)
+ 900 log(0.7788 - 1) - 1125 log(0.7788 + 1) + 800 sqrt(3) tan^(-1)((2 0.7788 +
1)/sqrt(3))))/3600
Input interpretation
185
From which:
((((166.722 -1011.57 i) - 570/(0.7788 - 1) + 30/(0.7788 - 1)^2 - 3475 log(1 – 0.7788)
+ 900 log(0.7788 - 1) - 1125 log(0.7788 + 1) + 800 sqrt(3) tan^(-1)((2 0.7788 +
1)/sqrt(3))))/3600 )-64/10^2
Input interpretation
Result
Polar coordinates
1.64623 ≈ ζ(2) = 𝜋2
6= 1.644934… (trace of the instanton shape)
Polar forms
191
From the sum of the first two mock 5.3425012228441…. subtracting the result of
the third mock 7.999997103998…. ≈ 8 R3 , we obtain:
(7.999997103998 - 5.3425012228441)-1.0018674362
where
Input interpretation
Result
1.6556284449539 result that is very near to the 14th root of the following
Ramanujan’s class invariant 𝑄 = 𝐺505/𝐺101/5 3 = 1164.2696 i.e. 1.65578...
Indeed, from:
113+5 505
8+ 105+5 505
8
314
= 1,65578…
192
And:
(7.999997103998 - 5.3425012228441)-(1/0.9568666373)
where
Input interpretation
Result
1.6124181649…. result that is a very good approximation to the value of the golden
ratio 1.618033988749...
And again:
1/2(((7.999997103998 - 5.3425012228441) -(((24*8-((8*2)-4)) π)/(24^2+1)))+
((7.999997103998 - 5.3425012228441)-(1/0.9568666373)))
where
193
Input interpretation
Result
1.6449339….≈ ζ(2) = 𝜋2
6= 1.644934… (trace of the instanton shape)
Possible closed forms
Alternative representations
195
Integral representations
(5.3425012228441 + 7.999997103998)^108 * (Catalan + 2 - π + π log(3/2))
Input interpretation
Result
0.35159968099…*10
122 ≈ ΛQ
The observed value of ρΛ or Λ today is precisely the classical dual of its quantum
precursor values ρQ , ΛQ in the quantum very early precursor vacuum UQ as
determined by our dual equations.
196
Fundamental are the following values: Λ = 2.846 * 10-122
that is the actual value of
the Cosmological Constant that is precisely, the classical dual of its quantum
precursor value ΛQ = 0.3516 * 10122
in the quantum very early precursor vacuum.
(New Quantum Structure of the Space-Time - Norma G. SANCHEZ - arXiv:1910.13382v1
[physics.gen-ph] 28 Oct 2019)
Alternative representations
Series representations