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$\det(A) = \det(P)\cdot \det(L)\cdot\det(U) = \det(P)\cdot \det(U),$

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

a b = e bloga

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).

Páginas

viernes, 28 de enero de 2011

prime numbers hypothesis riemman solved

Relation to the Riemann zeta function

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:

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Páginas

viernes, 28 de enero de 2011

prime numbers hypothesis riemman solved

Relation to the Riemann zeta function

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:

No hay comentarios:

Publicar un comentario

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: