viernes, 28 de enero de 2011

prime numbers hypothesis riemman solved











 \det(A) = \det(P)\cdot \det(L)\cdot\det(U) = \det(P)\cdot \det(U),
because det(L) = 1; the right hand side is easily computed as the product of all diagonal elements of U multiplied with the determinant of the permutation matrix P (which is +1 for an even permutation and is -1 for an odd permutation). This is more efficient than calculating the determinant of A because the determinant of a (upper or lower) triangular matrix is the product of its diagonal elements.
A small example:
A =
\begin{bmatrix}
           6 & 3 \\
           4 & 3 \\
        \end{bmatrix}
 PA = LU =
      \begin{bmatrix}
           1 & 0 \\
           0 & 1 \\
        \end{bmatrix}
\cdot
      \begin{bmatrix}
           6 & 3 \\
           4 & 3 \\
        \end{bmatrix}
=
      \begin{bmatrix}
           1 & 0 \\
           2/3 & 1 \\
        \end{bmatrix}
\cdot
        \begin{bmatrix}
           6 & 3 \\
           0 & 1 \\
        \end{bmatrix}
Therefore
 \det(A) = \det(P) \cdot \det(U) = 1\cdot 6 = 6.\
due the calculations are linnear
det(A-lambdasubi*I=C)=0->in a base B we obtain the diagonal matrix , 
B*C*B^-1=I
       (2  0  0  0.................................0)         (   )=(-(2^n)-1 n=0
       (0  4  0  0..................................0)       ----     (-2^n)-2=0
       (----------------------------...............0)        ...            ...
C=((0  0  2^(a-1)  00..,00.....................0)  X= (2^(n-1)-(2^n))=0=-1/2=lambdaN for all n
     (0  0      3      0    0    --...............0) 
      (   ----------------------------------------
      (0  0     0     3^4 0   ....................0)                                                                                                                                           
      (0   0   0      0     0      3^b 0..0..00)
      (   ........................................          )                                                                                                           
       (0......                ...pi^k................0)
   .    (..................................................0)
         (0....----------..........................pn^n)  , det(C)=p(n+1)-1 even,pair number=2^a*3^b*5^c*pn^n  ,
 it conforms the system of equations:     det(C-2^a*I)=0..det(C-3^b*I)=0

we also can use the conjecture of coldbach pi+pm=((p(n+1)-1))*(1/(p(n+1)-1)^n)=even
(1/(p(n+1)-1)^n)*(pi+pm)=((p(n+1)-1))->(1/(p(n+1)-1)^n)*(2^a*3^b*5^c*p(i-1)^(i-1)+2^a*3^b*5^c*p(m-1)^(m-1))
\zeta(s) = \prod_{p \text{ prime}} \frac{1}{1-p^{-s}}= \frac{1}{1-2^{-s}}\cdot\frac{1}{1-3^{-s}}\cdot\frac{1}{1-5^{-s}}\cdot\frac{1}{1-7^{-s}} \cdots \frac{1}{1-p^{-s}} \cdots
q,c,d
seeing at another perspective 
      (1 0 0 )            (0 0  1)
C=(2 0 0...) C-1=(0 0 1/2)
(0 4 0...)              (0 1/4 0)
(0 0 8...)              (1/8 0 0)
(.........  )               ............
(0 0 0 2^a)         (1/2^a 0 0)
Det[(C)tensor product x Ntimes x C x (D) .. Ntimes D x ... N x    Ntimes x..xN]
= ((lambda1=-1=(1/(2*2^2*2^4*2^3*..2^N)^(2*2^2*2^4*2^3*..2^N))+1)*(lambda2=
-1=1/(3^2*3^3*3^4*..*3^N)^ (3^2*3^3*3^4*..*3^N))+1)*...*(lambdaN=
-1=1/(pN*pN^2*pN^3*..*P^N)^(pN*pN^2*pN^3*..*P^N))+1)=zeta(S)=0


\sum_{n\ge 1} \frac{1}{n^s} = \prod_p \sum_{k\ge 0}(p^{-s})^k=\left(1 + \frac{1}{2^s} + \frac{1}{4^s} + \frac{1}{8^s} + \cdots \right) \left(1 + \frac{1}{3^s} + \frac{1}{9^s} + \frac{1}{27^s} + \cdots \right) \cdots \left(1 + \frac{1}{p^s} + \frac{1}{p^{2s}} + \frac{1}{p^{3s}} + \cdots \right) \cdots
\prod_p^\infty \left(1-\frac{1}{p^s}\right) = \left( \prod_p^\infty \frac{1}{1-p^{-s}} \right)^{-1} = \frac{1}{\zeta(s)}. =((p^S)-1)/p^S=
Det(C-1)*Det[(C-1)tensor product x (D-1)x..x(N-1)]*Det(1/C^t (tensor product=x) 1/D^t ) x 1/E^t  x..1/(N-1)^t)*
Det(CxDx..xN)*(C^-1xD^-1x..xN^-1)=Ix..Ntimes..xI lambdai=1^N just need to apply in a recurrent form conjecture of coldback prime1+prime2=even so we also can use the conjecture of coldbach pi+pm=((p(n+1)-1))*(1/(p(n+1)-1)^n)=even develop and obtain final soilution

Relation to the Riemann zeta function

Some derivatives of fractional harmonic numbers are given by:
\frac{d^n H_x}{dx^n} = (-1)^{n+1}n!\left[\zeta(n+1)-H_{x,n+1}\right]
\frac{d^n H_{x,2}}{dx^n} = (-1)^{n+1}(n+1)!\left[\zeta(n+2)-H_{x,n+2}\right]
\frac{d^n H_{x,3}}{dx^n} = (-1)^{n+1}\frac{1}{2}(n+2)!\left[\zeta(n+3)-H_{x,n+3}\right]
And using Maclaurin series, we have for x<1 :
H_x = \sum_{n=1}^{\infin}(-1)^{n+1}x^n\zeta(n+1)
H_{x,2} = \sum_{n=1}^{\infin}(-1)^{n+1}(n+1)x^n\zeta(n+2)
H_{x,3} = \frac{1}{2}\sum_{n=1}^{\infin}(-1)^{n+1}(n+1)(n+2)x^n\zeta(n+3)
where ζ(n) is the Riemann zeta function.


C=(C/2-2*lambda*IxIxI..I)             ->lambdasubi=2^i
D=(3 0 0...)
     (0  9 0...)
     (0  0 27..)
     .....
     (0  0  0  3^b)
D=D/3-3lambdaIxIxI..I   .>lambdasubj=3^j
E=(5 0 0...0)
     (0 0 25..0)
      ...........
      (0 0 0 5^c)
....
N=(pn 0 0 ..0)
     (0  pn^2 0..0)
 .   ................
     (0   0      0 pn^z)


|((C tensor product D tensor product E ... tensor product N)-(lambda=(2^a*3^b*5^c*p(i-1)^(i-1) ( I tensor product I ... tensor product I) ....|=0->lambdasubi=pi= 
PA=LU->PA=I*B
B*C=I*B->C=B^-1*B=B^t*B/det|B|->det(C) normalized= p(n+1)-1=(2^a/(2^a*3^b*5^c*pn^n))*(3^b/(2^a*3^b*5^c*pn^n))*..*((pn^n/(2^a*3^b*5^c*pn^n))=
(p(n+1)-1)=((p(n+1)-1))*(1/(p(n+1)-1)^n)
det(A-lambdasubi*I=C)=0

\det(\mathbf{AB}) = \det(\mathbf{A})\cdot\det(\mathbf{B})
               
Method to obtain prime numbers:
det|A-lambdasubi*I|=0
A=(1 111111111111111111..
     (2 4 8 16 32 64......
     (3 9 27 71..........................
     (4 16 24...
     (5 25 125..
    (6 36  36^2......
    (7  14  98 .....
    (8  64..           mxm
    ....
   ...
det|(A-lambdasubi*I)tensorprod.(A-lambdasubj*I)
..tensorprod(A-lambdasubk*I)|=0

And now the fast sieve of Arostothenes.
A in binary

A=(01 01  01  01 01 01 01 01 01 01 01 01...
    *(10 100 1000 10000 100000 1000000......

     (11 111 (in decimal 27 81..........................
    *(100 10000 1000000...
     (101 (in decimal..25 125..
    ((in decimal 6 36  36^2......
    ((in decimal7  14  98 .....
    *(1000  1000000000..           mxm

for base=3*2=6 6 36 36^2 36^3=
    *(6|base3*2 10 100 1000 10000 100000 1000000......

another interesting observation:
pN/((pN)-1)->Dividendus=quotient*divisor+rest->
pN=((pN)-1)*1+1
P(N)-1/(P(N-1)-1=2^a
P(N)-1/(P(N-1)-1=2^a due both are even
P(N)/P(N)-1/P(N-1)/P(N-1)-1=
=(P(N-1)-1)*P(N)/(P(N)-1)*P(N-1)=1=(1/2^a)*P(N)/(P(N)-1) =root(N for each N->0 to inf)((1/2^a)*P(N)/(P(N)-1))

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------MOST INTUITIVE SOLUTION-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------


because det(L) = 1; the right hand side is easily computed as the product of all diagonal elements of U multiplied with the determinant of the permutation matrix P (which is +1 for an even permutation and is -1 for an odd permutation). This is more efficient than calculating the determinant of A because the determinant of a (upper or lower) triangular matrix is the product of its diagonal elements.
A small example:


Therefore


due the calculations are linnear
det(A-lambdasubi*I=C)=0->in a base B we obtain the diagonal matrix , 
B*C*B^-1=I
       (2  0  0  0.................................0)         (   )=(-(2^n)-1 n=0
       (0  4  0  0..................................0)       ----     (-2^n)-2=0
       (----------------------------...............0)        ...            ...
C=((0  0  2^(a-1)  00..,00.....................0)  X= (2^(n-1)-(2^n))=0=-1/2=lambdaN for all n
     (0  0      3      0    0    --...............0) 
      (   ----------------------------------------
      (0  0     0     3^4 0   ....................0)                                                                                                                                           
      (0   0   0      0     0      3^b 0..0..00)
      (   ........................................          )                                                                                                          
       (0......                ...pi^k................0)
   .    (..................................................0)
         (0....----------..........................pn^n)  , det(C)=p(n+1)-1 even,pair number=2^a*3^b*5^c*pn^n  ,
 it conforms the system of equations:     det(C-2^a*I)=0..det(C-3^b*I)=0


we also can use the conjecture of coldbach pi+pm=((p(n+1)-1))*(1/(p(n+1)-1)^n)=even
(1/(p(n+1)-1)^n)*(pi+pm)=((p(n+1)-1))->(1/(p(n+1)-1)^n)*(2^a*3^b*5^c*p(i-1)^(i-1)+2^a*3^b*5^c*p(m-1)^(m-1))


q,c,d
seeing at another perspective 
      (1 0 0 )            (0 0  1)
C=(2 0 0...) C-1=(0 0 1/2)
(0 4 0...)              (0 1/4 0)
(0 0 8...)              (1/8 0 0)
(.........  )               ............
(0 0 0 2^a)         (1/2^a 0 0)
Det[(C)tensor product x Ntimes x C x (D) .. Ntimes D x ... N x    Ntimes x..xN]
= ((lambda1=-1=(1/(2*2^2*2^4*2^3*..2^N)^(2*2^2*2^4*2^3*..2^N))+1)*(lambda2=
-1=1/(3^2*3^3*3^4*..*3^N)^ (3^2*3^3*3^4*..*3^N))+1)*...*(lambdaN=
-1=1/(pN*pN^2*pN^3*..*P^N)^(pN*pN^2*pN^3*..*P^N))+1)=zeta(S)=0




=((p^S)-1)/p^S=
Det(C-1)*Det[(C-1)tensor product x (D-1)x..x(N-1)]*Det(1/C^t (tensor product=x) 1/D^t ) x 1/E^t  x..1/(N-1)^t)*
Det(CxDx..xN)*(C^-1xD^-1x..xN^-1)=Ix..Ntimes..xI lambdai=1^N just need to apply in a recurrent form conjecture of coldback prime1+prime2=even so we also can use the conjecture of coldbach pi+pm=((p(n+1)-1))*(1/(p(n+1)-1)^n)=even develop and obtain final soilution
Relation to the Riemann zeta function
Some derivatives of fractional harmonic numbers are given by:






And using Maclaurin series, we have for x<1 :






where ζ(n) is the Riemann zeta function.


C=(C/2-2*lambda*IxIxI..I)             ->lambdasubi=2^i
D=(3 0 0...)
     (0  9 0...)
     (0  0 27..)
     .....
     (0  0  0  3^b)
D=D/3-3lambdaIxIxI..I   .>lambdasubj=3^j
E=(5 0 0...0)
     (0 0 25..0)
      ...........
      (0 0 0 5^c)
....
N=(pn 0 0 ..0)
     (0  pn^2 0..0)
 .   ................
     (0   0      0 pn^z)


|((C tensor product D tensor product E ... tensor product N)-(lambda=(2^a*3^b*5^c*p(i-1)^(i-1) ( I tensor product I ... tensor product I) ....|=0->lambdasubi=pi= 
PA=LU->PA=I*B
B*C=I*B->C=B^-1*B=B^t*B/det|B|->det(C) normalized= p(n+1)-1=(2^a/(2^a*3^b*5^c*pn^n))*(3^b/(2^a*3^b*5^c*pn^n))*..*((pn^n/(2^a*3^b*5^c*pn^n))=
(p(n+1)-1)=((p(n+1)-1))*(1/(p(n+1)-1)^n)
det(A-lambdasubi*I=C)=0




               


Method to obtain prime numbers:
det|A'-lambdasubi*I|=0
det |M'|<>0,M' = operaciones lineales quitando los dependientes con A'
A'=(1 111111111111111111..
     (2 4 8 16 32 64......
     (3 9 27 71..........................
     (4 16 24...
     (5 25 125..
    (6 36  36^2......
    (7  14  98 .....
    (8  64..           mxm
    ....
A=(1 0 0 0  0 0 00 0 0 0 0..
     (2 4 0 0  0 0 00  0 0 0 ......
     (3 9 27 0 0  0 0 0 0 0 0..........................
     (4 16 32 64 0 0 0  00 0...
     (5 25 125 625 5^6 0 0 0..
    (6 36  36^2 36^4 36^5 36^6 36^7......
    ( .......................................    mxm
det|A-lambdasubi*I|=0
hacemos combinaciones lineales de las filas que nos da la misma det de la matriz
M'=(
(fila1=(pot2)A0,0..n+A1,0..n+A3,0..n+A7,0..n+A(8+64-1),0..n...
(fila2(pot3)A0,0..n+A2,0..n+A8,0..n+A16,0..n+A24,0..n+......
(fila3=(pot5).A0,0..n+A4,0..n..
tenemos que , we have that
fila1=2^m=Am,0=pow(2m)(A0,0),A7,0=pow(2)(A3,0);A3,0=pow(2)(A1,0=2*A0,0))
fila2=3^m=Am,0=pow(3m)(A0,0),A8,0=pow(3)(A4,0);A3,0=pow(3)(A2,0=3*A0,0))
fila3=5^m=Am,0=pow(5m)(A0,0),A100,0=pow(5)(A20,0);A20,0=pow(3)(A4,0=3*A0,0))
luego then
fila1=2^m=Am,0;xm*pow(2m)(A0,0)+..+x2*A7,0+x1*pow(2)(A3,0);A3,0+x0*pow(2)(A1,0=2*A0,0))=0
fila2=3^m=Am,0;xm*pow(3m)(A0,0)+..+x2*A8,0+x1*pow(3)(A4,0);A3,0+x0*pow(3)(A2,0=3*A0,0))=0
fila3=5^m=Am,0;xm*pow(5m)(A0,0)+..+x2*A100,0+x1*pow(5)(A20,0);A20,0+x0*pow(5)(A4,0=3*A0,0))=0
luego then
det|fila1=2^m=(Am,0)pow(2m)-lambdam..  ,,A7,0 ...pow(2)(A3,0);A3,0.... ... pow(2)(A1,0=2*A0,0))=0
|fila2=3^m=Am,0;pow(3m)(A0,0).......x2*A8,0;pow(3)(A4,0)-lambda1;A3,0*pow(3)(A2,0=3*A0,0))=0
|fila3=5^m=Am,0;xm*pow(5m)(A0,0)+..+x2*A100,0+x1*pow(5)(A20,0);A20,0+pow(5)(A4,0=3*A0,0))-lambda0=0
(1-lamdba1)*(2-lambda2)*(3-lambda3)*..*(p-lambdap)=0
(1-lamdba1)^-S*(2-lambda2)^-S*(3-lambda3)^-S*..*(p-lambdap)^-S=0^-S=0
that is the case the matrix M' has 2 rows repeated -S/2=a, -S/3=b,.. integer that gives out trivial zeroes
for det(A)=0-> prod(Ai,i)*...*((Ai-m,i-m)^-S=Aii^(-m*S)) S=-2,-3,-5,-7..prime gives out trivial zeroes
so probe the hipothesis of riemman results trivial:
((Ai-m,i-m)^-S=Aii^(m/2)) that is the definition of the trigonal matrix S=-1/2 S=-1/2+ib not altering the condition of det(A)=0



prod(Ai,i)*...*((Ai-m,i-m)^-S=Aii^(-m*S))
thats it the polynomia 



ab = ebloga



0=(number-lambda)^-S=number^m-lambda^-mS'
Ai+i=Ai+i,0 if(Ai-1,0=pi-1->A(i+i)-1,0->p(i+i-1)
A=(1 0 0 0
    (2 4  0 0
    (3 9 27 0
    (4 16 32 64 0

The simplest method of computing an requires n−1 multiplication operations, but it can be computed more efficiently as illustrated by the following example. To compute 2100, note that 100 = 64 + 32 + 4. Compute the following in order:

22 = 4
(22)2 = 24 = 16
(24)2 = 28 = 256
(28)2 = 216 = 65,536
(216)2 = 232 = 4,294,967,296
(232)2 = 264 = 18,446,744,073,709,551,616
264 232 24 = 2100 = 1,267,650,600,228,229,401,496,703,205,376
This series of steps only requires 8 multiplication operations instead of 99 (since the last product above takes 2 multiplications).

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