2011年10月19日水曜日

Turing Machine

I read "BEYOND REASON".
I don't know about Turing Machine, but this is quite interesting because I can still find the similarity with web programing in these days. I also count prime numbers, so I can study from the algorithm of Turing Machine. The realm is completely different. However, it seems to be a new light for me.

Turing Machine is like an infinite tape. It read a sign from each cell.
For example, there are three cells, such as 0(zero),1,X.





0

1

X

A

0NA

1NA

XRB

B

0RB

1RB

0RC

C

XLD

XLD

XLD

D

0LD

1LD

XNE

E

Stop

Stop

Stop


This is like a dictionary or code.
0="Put 0 in the cell". 1="Put 1 in the cell". N="Don't move the cell". A="Stay in the same cell". X="Put X in the cell". R="Move the cell to the right". B="The new state". C="The end of the line.Then put X". L="Move the cell to the left". D="The new state". E="The end".

When you see A, there are three cells such as 0=(0NA), 1=(1NA), X=(XRB). 0NA means that put 0 in the cell and don't do anything. 1NA means that put 1 in the cell and don't do anything. XRB means that put X in the cell and move it to the right for a new state.
After you put X in a cell, you can go to B.
In B, there are 0=(0RB) 1=(1RB) X=(0RC).
You can adjust the same pattern from the dictionary, then you see that C means the end. However, you put X, so the line don't stop. It goes to C.
In C, it moves to the left which means the reverse.
Finally, you see the end in D,X=(XNE).
However, if you don't put XNE and you increase cells like D1,D2,D3・・・, Turing Machine may never stop for ever.

2011年8月4日木曜日

J function

When you see the golden key, π(x) would be related to zeros of Riemann's zeta function. J function is also based on π(x). Therefore, if you counted prime numbers, you would guess zeros of Riemann's zeta function.

ζ(s)=1/(1-1/2^s)×1/(1-1/3^s)×1/(1-1/5^s)×・・・

When s is 0.5+zi, ζ(s)=0.
Here is a Riemann's zero.
s=14.1435657,21.0279853,・・・

In order to use the calculator, you need to put an integer which must be close to 0.5+zi. It is very hard to guess it, because there would be no pattern about it. However, when you calculate J function, you would know that it is almost the same with 0.5+zi. Of course, this isn't perfect. J function is related to prime numbers. The pattern of prime numbers would be fractal, and I can calculate π(x). Therefore, as long as zeros of Riemann's zeta function is connected to π(x), it would move with the same pattern as inflation. I know that the calculation is complicated.

x→J(x)→0.5+zi
40→14.61667→14.1435657
65→21.5333→21.0279853
80→25.28333→25.01585333
105→30.5333→30.42810832
110→32.5333→32.93879514
137→37.5095→37.58932556
153→40.5095→40.92105478





Moreover, when you calculate π(45+10x), 0.5+zi is zeros of Riemann's zeta function and π(45+10x) is almost close to z. The differences between π(45+10x) and z are around 2000?.


2011年7月5日火曜日

Cantor set

I don't understand what infinity is, but Cantor set is almost similar with prime numbers and my blog.

At first, you get rid of the center of each block.



This is the infinite pattern, so I can also write this. You keep deleting the center.


F0=1
F1=1/3
F2=1/9
F3=1/27



F(x)=1/3^x



If x is infinite, F(x) may be Zero, but it isn't.
Imaginary lines may be real, if zero exists.
Basically, Cantor set is expressed by on and off, but zero would mean none duality of duality.


You always have the one block, and you take the center off.

Therefore, 0*3^∞ must be 1, and this is ridiculous.
Is this the philosophical problem?
I'm not so curious about it.


2011年5月1日日曜日

Fractal

I found the pattern of prime numbers, but I know that it is almost the same with sieve of Eratosthenes.

When you see odd numbers, there is symmetry and you can find sieve of Eratosthenes.




3

5

7

9

11

3

9

15

21

27

33

5

15

25

35

45

55

7

21

35

49

63

77

9

27

45

63

81

99

11

33

55

77

99

121

I mean x=(3,5,7,9,11) y=(3,5,7,9,11).
Therefore, you can see z=x*y and z=x+Σ2x or z=y+Σ2y. 9=3*3 and 9=3+2*3,15=3*5 and 15=3+4*3 or 15=5*3 and 15=5+2*5, 21=3*7 and 21=3+6*3 or 21=7*3 and 21=7+2*7・・・・.
z is not prime numbers, so x'=(3,5,7,11) y'=(3,5,7,11).

However, when you add 2, the symmetry is broken.



3

5

7

9

11

2

6

10

14

18

22

3

9

15

21

27

33

5

15

25

35

45

55

7

21

35

49

63

77

9

27

45

63

81

99

11

33

55

77

99

121

x=(3,5,7,9,11) y=(2,3,5,7,9,11) z=x*y and z=x+Σ2x or z=y+Σ2y
x' and y' are prime numbers. x'=(3,5,7,11) y'=(2,3,5,7,11)

Therefore,

when you can see z=x*y and z=x+Σ2x or z=y+Σ2y, z is not prime numbers.


This would be the infinite pattern.

2011年1月22日土曜日

The Golden Key

Riemann's zeta function is known as contraction.

This is also contraction.

Then you can get the golden key.

Zeros of Riemann's zeta function is related to π(x).

Therefore, if you can know the golden key, you may understand zeros of Riemann's zeta function.

As long as I use PC to count prime numbers, it seems to be the same with J(x). This is assumption. I know what I don't know.

2011年1月17日月曜日

A part is everything

It is very difficult to see what Riemann's zeta function is. I don't write about 0 and I can't understand it. However, it is connected with prime numbers.

Now I got this formula.
logζ(s)=1/2^s+(1/2*1/2^2s)+(1/3*1/2^3s)+・・+1/3^s+(1/2*1/3^2s)+(1/3*1/3^3s)+・・1/5^s+(1/2*1/5^2s)+(1/3*1/5^3s)+・・・

Then, you pick up (1/2*1/3^2s) from it. I don't know the reason, but I follow Prime obsession.


I can't understand this process, but this leads to J function


You may see a form which is almost a rectangular (width="3^2→∞", height="1/2").
J(x) include ∞, and 1/2*s*∫[3^2→∞]x^(-s-1)dx is also ∞.
Is this like a black hole?



2011年1月16日日曜日

Riemann's zeta function

Riemann's zeta function is expressed like this. This depends on Sieve of Eratosthenes.



According to index principle,

log1/A=-logA,so

Isaac Newton found

x=1/p^s (p is prime)




This pattern is really complicated and miraculous. I'm in the spiral road.
Then, you need infinitesimal calculus.

I referred to Prime obsession.

2011年1月15日土曜日

Power Law

I counted prime numbers up to 100000000 by using visual basic. The graph is like this.Prime numbers increase infinitely.










However, when you add more integral numbers, there are less prime numbers. This is known as power law.





2011年1月14日金曜日

Inflation

I programmed to count prime numbers by PC, so I can see the same pattern from "Prime Obsession". There is the function about prime numbers.


π means the function of prime numbers up to x.
For example, you can put 100 to x.
10 is the square root of 100, 4.6415・・is the cube root of 100, 3.1622・・is the fourth root of 100・・,2.1544・・is the sixth root of 100,and 1.9306・・is the seventh root of 100. 2 is the first prime number,so you can ignore the seventh root of 100.

Therefore
J(100)=π(100)+1/2π(10)+1/3π(4.64・・)+1/4π(3.16・・)+1/5π(2.51・・)+1/6π(2.15・・)+0+0・・・

You can count prime numbers by 100, so you see π(100)=25, π(10)=4, π(4.64・・)=2, π(3.16・・)=2, π(2.51・・)=1, π(2.15・・)=1.

J(100)=25+(1/2×4)+(1/3×2)+(1/4×2)+(1/5×1)+(1/6×1)

This is the graph.


X is a prime number.
When x is 2, J(2)=1. When x is 3, J(3) jump to 2. When x is 5, J(5) jump to 3.5.

This graph would go up infinitely.

2011年1月9日日曜日

The pattern of prime numbers

I found the pattern of prime numbers. It must be ridiculous but it seems to be almost perfect as long as I know. The realm includes infinity, so it is impossible for me to prove it. However, it keeps moving with the same pattern.
You may say that it must be a fractal which means that a part includes everything. If you are interested in a long spiral road, read this blog. I will extend a stupid story from prime numbers to the universe.

ρ must be optional odd numbers.
Ш is apparent prime numbers such as 2,3,5.
In this case, you can ignore 2 because only odd numbers must be the target to expand prime numbers.



If ρ is divided by Ш and you can get the integral answer, it is Шn^2. It must be Composite numbers before Шn^2 such as 15=5*3.


These are not prime numbers.



For example, you can find prime numbers up to 100 by following this theory.
《3、5、7、9、11、13,15、17、19、21、23、25、27、29、31、33、35、37、39、41、43、45、47、49、51、53、55、57、59、61、63、65、67、69、71、73、75、77、79、81、83、85、87、89、91、93、95、97、99》is the odd numbers.

3,5,7 must be the apparent prime numbers because of √100=10.

Ш1=3,Ш2=5,Ш3=7

9 is divided by Ш1=3, so ρ=9 and 9+2(д-1)Ш1 is working.

д>1

Therefore
9+2×3=15、9+4×3=21、9+6×3=27、9+8×3=33、9+10×3=39、9+12×3=45、9+14×3=51、9+16×3=57、9+18×3=63、9+20×3=69、9+22×3=75、9+24×3=81、9+26×3=87、9+28×3=93、9+30×3=99

15、21、27、33、39、45、51、57、63、69、75、81、87、93、99 are not prime numbers.

The next is Ш2=5 and 25 is divided by this number.
ρ=25 and 25+2(д-1)Ш2 is working.
Therefore
25+2×5=35、25+4×5=45、25+6×5=55、25+8×5=65、25+10×5=75、25+12×5=85、25+14×5=95

35、45、55、65、75、85、95 are not prime numbers.

The next is Ш3=7 and 49 is divided by this number.
ρ=49 and 49+2(д-1)Ш3 is working.
Therefore
49+2×7=63、49+4×7=77、49+6×7=91

63、77、91 are not prime numbers.

Finally you can find prime numbers by 100.

3、5、7、11、13、17、19、23、29、31、37、41、43、47、53、59、61、67、71、73、79、83、89、97

You can't break this pattern until the end, and you never know.
Шn+1 must be expansion of Шn^2+2(д-1)Шn.


C=2(д-1)Шn



This would be fractal.
Moreover, this is expressed by squares, so you can say C=αi^2 (i is an imaginary number).
For example, Ш2=5=3^2+4i^2=9-4 and Ш3=7=5^2+18i^2=25-18.
There is no pattern in C, but it is based on prime numbers.



Moreover, Шn must be P.
Here is Riemann's zeta function.



I exclude 2, but ζ(s) is still fractal because of multiplication.