In binary representation of
2
are there more ones or zeros?
​
2
is an irrational number. Regardless what base numbering system youre using. There is no way to tell how many of each there are... because they go on forever.
The first so many 1.01101010000010... no predictability, there shouldn’t be any “pattern” to the binary representation of it, and therefore, statistically, the ratio of the numbers of 1s and 0s will approach 1.
Is there a pattern?,So there is, and we will try to discover it.
How to find? well:​
Using this identity:
Solving sum :
In[]:=
Floor[x]==x-1/2+1/Pi*Sum[Sin[2*k*Pi*x]/k,{k,1,Infinity}]
Out[]=
⌊x⌋x+
log(1-
2πx

)-log(
-2πx

(-1+
2πx

))
2π
-
1
2
At the first I’m find closed expression integer sequences of [OEIS A084188 ][2] and substitute
x=
n+
1
2
2
to first equation.
In[]:=
Floor[x]==-(1/2)+x+(I(Log[1-E^(2Iπx)]-Log[E^(-2Iπx)(-1+E^(2Iπx))]))/(2π)/.x->(Sqrt[2]*2^n)
Out[]=

n+
1
2
2

n+
1
2
2
+
log1-
π
n+
3
2
2

-log
-π
n+
3
2
2

-1+
π
n+
3
2
2


2π
-
1
2
a checks if works:
In[]:=
Table[-(1/2)+2^(1/2+n)+(I(Log[1-E^(I2^(3/2+n)π)]-Log[E^(-I2^(3/2+n)π)(-1+E^(I2^(3/2+n)π))]))/(2π)/.n->m,{m,0,19}]//N//Chop//Rationalize
Out[]=
{1,2,5,11,22,45,90,181,362,724,1448,2896,5792,11585,23170,46340,92681,185363,370727,741455}
then I’m need solve for
b(n)
from equation
a(0)=1
,
a(n+1)=2a(n)+b(n+2)
In[]:=
sol=RSolve[{b[n+2]==a[n+1]-2*a[n],a[n]==-(1/2)+2^(1/2+n)+(I(-Log[1-E^(-I2^(3/2+n)π)]+Log[1-E^(I2^(3/2+n)π)]))/(2π),b[0]==1,b[1]==0},{b[n],a[n]},n]​​sol2=FullSimplify[b[n]/.sol[[1]],Assumptions->{n>0,n∈Integers}]
We simplify
B
n
expression to very simple form:
In[]:=
sol22=1/2-(2ArcTan[Cot[2^(-(3/2)+n)π]])/π+ArcTan[Cot[2^(-(1/2)+n)π]]/π
This is closed formula for the n-th digit in the binary representation of
2
for
n>0
and
n∈
In[]:=
Table[sol22/.n->m,{m,1,50}]//Expand
Out[]=
{1,0,1,1,0,1,0,1,0,0,0,0,0,1,0,0,1,1,1,1,0,0,1,1,0,0,1,1,0,0,1,1,1,1,1,1,1,0,0,1,1,1,0,1,1,1,1,0,0,1}
checks if works:
In[]:=
mm=N[Sqrt[2],16];RealDigits[mm,2][[1]]
Out[]=
{1,0,1,1,0,1,0,1,0,0,0,0,0,1,0,0,1,1,1,1,0,0,1,1,0,0,1,1,0,0,1,1,1,1,1,1,1,0,0,1,1,1,0,1,1,1,1,0,0,1,1,0,1}
this last integer sequences can be found here OEIS A004539.
Another test checking closed expression to numerical value for the square root of two of base-10 :
In[]:=
sol3=Table[Block[{$MaxExtraPrecision=10000},N[Limit[sol22,n->m],10]],{m,1,600}]//Rationalize//Quiet​​N[FromDigits[{Drop[sol3,0],1},2],166]
Out[]=
1.414213562373095048801688724209698078569671875376948073176679737990732478462107038850387534327641572735013846230912297024924836055850737212644121497099935831413222666
In[]:=
N[Sqrt[2],166]
Out[]=
1.414213562373095048801688724209698078569671875376948073176679737990732478462107038850387534327641572735013846230912297024924836055850737212644121497099935831413222666
At the end of post a interesting sum:
In[]:=
N[2*Sum[(1/2-(2ArcTan[Cot[2^(-(3/2)+n)π]])/π+ArcTan[Cot[2^(-(1/2)+n)π]]/π)*2^(-n),{n,1,164}],50]
Out[]=
1.4142135623730950488016887242096980785696718753769

Appendix: To simplify
B
n
formula.

We have:
In[]:=
sol2=(I(-Iπ+2Log[1-E^(-I2^(-(1/2)+n)π)]-2Log[1-E^(I2^(-(1/2)+n)π)]-Log[1-E^(-I2^(1/2+n)π)]+Log[1-E^(I2^(1/2+n)π)]))/(2π)​​sol5=FullSimplify[Re[sol2]//ComplexExpand,Assumptions->n∈Integers]​​Table[sol5,{n,1,5}]
Out[]=

π-4
-1
tan
sin(
2
π)
1-cos(
2
π)
+2
-1
tan
sin(2
2
π)
1-cos(2
2
π)
2π
,
π-4
-1
tan
sin(2
2
π)
1-cos(2
2
π)
+2
-1
tan
sin(4
2
π)
1-cos(4
2
π)
2π
,
π-4
-1
tan
sin(4
2
π)
1-cos(4
2
π)
+2
-1
tan
sin(8
2
π)
1-cos(8
2
π)
2π
,
π-4
-1
tan
sin(8
2
π)
1-cos(8
2
π)
+2
-1
tan
sin(16
2
π)
1-cos(16
2
π)
2π
,
π-4
-1
tan
sin(16
2
π)
1-cos(16
2
π)
+2
-1
tan
sin(32
2
π)
1-cos(32
2
π)
2π

Now we need find a pattern.The first ingredient in the formula:
We have:
In[]:=
FindSequenceFunction[{1,2,4,8,16},n]
Out[]=
-1+n
2
Finding the second ingredient in the formula is the same how I found at first:
In[]:=
FindSequenceFunction[{2,4,8,16,32},n]
Out[]=
n
2
Substitution:
In[]:=
FullSimplify[(π-(4ArcTan[Sin[nSqrt[2]π]/(1-Cos[nSqrt[2]π])]/.n->2^(n-1))+(2ArcTan[Sin[nSqrt[2]π]/(1-Cos[nSqrt[2]π])]/.n->2^n))/(2π)]//Expand
Out[]=
-1
tan
cotπ
n-
1
2
2

π
-
2
-1
tan
cotπ
n-
3
2
2

π
+
1
2

CITE THIS NOTEBOOK

Formula for computing sqrt(2) of binary numbers​
by Mariusz Iwaniuk​
Wolfram Community, STAFF PICKS, 2016
​https://community.wolfram.com/groups/-/m/t/1063480