P^2+Σ2P

When you keep adding integers infinitely, O (not Zero) is usable to express this difficulty. It is called Landau symbol. O(n), O(n^2), O(nlogn) This is expansion anyway. Traveling salesman problem is well known as Landau symbol. In this case, there are 4 places. The salesman starts from A and comes back to the same place. He must choose the fastest. This is easy one. However, when you keep adding places, the salesman would be confused. It is said that there are (n-1)!/2 choices. If you have 10 places, (10-1)!/2=(9*8*7*6*5*4*3*2*1)/2=181440. There are 181440 choices. If you have 14 places, (14-1)!/2=3113510400. This is expansion. Max=T(1) k=2 T(k)＞T(1) Max=T(k) k＜n and T(k+1)=T(n)＞T(k) Max=T(k+1)=T(n) ∴

Twin prime seems to be related to Galois theory. Galois described field extension and intermediate field. F=coefficient, K=field extension, M=intermediate field. G=Galois group. Therefore F⊂M⊂K G(K/M) G(K/F) ∴ G(K/M)⊂G(K/F) There is cubic equation. x^3+ax+b=(x-α)(x-β)(x-γ)=0 α+β+γ=0, αβ+βγ+αγ=a, αβγ=-b G e f1 f2 g1 g2 g3 α α β γ β α γ β β γ α α γ β γ γ α β γ β α e⊂H⊂G H1=(e,g1) H2=(e,g2) H3=(e,g3) H=(e,f1,f2) H e f1 f2 α α β γ β β γ α γ γ α β I define F⊂M⊂K and G(K/M)⊂G(K/F) Therefore, G=F and H=M. K⊂M⊂F This is upside-down. ∴ K(G(M))=M, G(K(H))=H

I describe symmetry which is based on Galois theory. I think that it must be related to prime numbers. Quadratic equation is like a mirror. There is the formula. f(x)=0 have two solutions, α and β. ∴α+β=-a As you know, f(α)=f(β)=0. If f(α)=β and f(β)=α, f(f(α))=α and f(β)=α. I define α+β=-a. Therefore f(α+β)=f(-a) f(α)+f(β)=β+α=-a β+f(β)=-a ∴f(β)=-a-β=α+β-β=α. There is symmetry.

When you count prime numbers in each 1000000, there are less prime numbers. This is known as power law . I have found the pattern of prime numbers . Шn+1=Шn^2+2(д-1)Шn This is absolute convergence . Therefore, it would be replaced as f(z). I can pick up all prime numbers infinitely, so, in this case, f(z) must be the sum of prime numbers in each 100. Z=Z(T)=x(T)+iy(T) T is a parameter. f(Z)=f(Z(T)) C is the curve of this graph, and you can divide it by N. T0=Ta,T1,T2,.....,TN=Tb Z0=Za,Z1,Z2,.....,ZN=Zb ΔZn≡Zn-Zn-1, ΔTn≡Tn-Tn-1 Z=Z(T) and ΔZn≡Zn-Zn-1 ∴

This figure is piling squares which are prime numbers. The pattern includes complex plane , so I think that I can describe Riemann surface. θ=2nπ 　(n=0,±1,±2........) and arg Z=θ+2nπ (-π＜θ≦π). Therefore, arg Z change according to n. n=0 →　-π＜arg Z≦π n=1 →　π＜arg Z≦3π n=5 →　9π＜arg Z≦11π n=-5 → -11π＜arg Z≦-9π ∴(2n-1)π＜arg Z≦(2n+1)π Complex planes Z depend on n, and these are piled infinitely.

I define how to expand circles in complex plane. Then, this is apparent. (n=0,±1,±2.......) Therefore －Π＜θ≦Π Z keep circling because of Z=X+iY

It is impossible to find all prime numbers, so Maclaurin's expansion is a good example to show convergence. If x had a certain value, this mathematical formula would be convergence. Then, you see complex number Z. This complex plane don't include point at infinity, so it must be absolute convergence . Z=X0 means expansion of the circle, and X0 is a radius of each circles.