How I found 2-≃0.00,000,000,02and +-2≃0.00,000,000,000,000,02
13
3CMRBE
13
3CMRB
10
-1
cos
λ
10
7561π
Found that Pi=.Then examined .Found it is approximately .Then guided Mathematica into finding out that ≃2, i.e.,≃1.9999999997462251.Found 1.9999999997462251≃≃2-.From of those approximations I found: ≃.00000000000000000000502 where r= = and c==t.
6ζ(2)
6
MRB
21/26
2
13/8
(3CMRBE)
13
3CMRBE
2-
5
ℋ
10
7561
10
-1
cos
λ
10
7561π
acombination
3(CMRB+r)-
8/13
2
r
π
1327c+6
20(187c+4)
∞
∫
0
Log[1+t]
-1+
t
In[]:=
3.141592653589
In[]:=
MRB constnat
In[]:=
CMRB=0.187859642462067120248517934054273230055903094900138786172004684
Out[]=
0.18785964246206712024851793405427323005590309490013878617200468
In[]:=
Sqrt[6*CMRB]
Out[]=
1.06167690696011784425556739456273027970600503880017403104590761
In[]:=
1.06167690696
In[]:=
6CMRB-2^(21/13)/E
Out[]=
-8.801364971397235293683902073831910352859892318361586×
-11
10
In[]:=
3CMRB-2^(8/13)/E
Out[]=
-4.400682485698617646841951036915955176429946159180793×
-11
10
In[]:=
N[2^(8/13)/(3E),20]
Out[]=
0.18785964247673606187
In[]:=
CMRB-2^(8/13)/(3E)
Out[]=
-1.466894161899539215613983678971985058809982053060264×
-11
10
In[]:=
13/8
(3CMRBE)
Out[]=
1.9999999997462251091717290341558057223094750587247134808008177
In[]:=
13
3CMRBE
In[]:=
1.9999999997462251
There is the possibility of being too greedy for precision. I wonder I was so in if the following.
There is the possibility of being too greedy for precision. I wonder I was so in if the following.
In[]:=
0.3495591344299050639041087
Here, I recognize:
Here, I recognize: