ComputerUser

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@vv 

i understand now,

by the way,

do you know how to predict a future matrix 

belong to which one of 3 matrix?

actually eigenvector has very small difference in 

Decimal value , can linear regression help to predict the exact eigenvector values?

i find book mention magnetron , neucleon these words, but I use Time series only to find Pauli equation. Is it the end after find Pauli equation ?

how to use Pauli equation for Time series?

@asa12 

i succeed to do and completed today

@Kitonum 

@Kitonum 

 

with(LinearAlgebra):
A := Matrix(2, 2, {(1, 1) = (1/6)*sqrt(3)+(1/2)*I, (1, 2) = (1/6)*sqrt(3)-(1/2)*I, (2, 1) = (1/6)*sqrt(3)-(1/2)*I, (2, 2) = (1/6)*sqrt(3)+(1/2)*I}):
with(MTM):
factor(Re(A));
A3 := Matrix([[0,0],[0,0]]):
for i from 1 to 2 do
 for j from 1 to 2 do
  A3[i,j] := Im(A[i,j]);
 od:
od:
counterd := 0:
for i from 1 to 2 do
 for j from 1 to 2 do
  if A2[i,j] mod ((1/6)*sqrt(3)) = True then
   counterd := counterd + 1:
  end if:
 od:
od:
counterd;
if counterd = 4 then
 for i from 1 to 2 do
 for j from 1 to 2 do
  A2[i,j] := A2[i,j]/((1/6)*sqrt(3)):
 od:
od:
end if:
A2;
A2 := Matrix([[0,0],[0,0]]):
for i from 1 to 2 do
 for j from 1 to 2 do
  A2[i,j] := Im(A[i,j]);
 od:
od:
counterd := 0:
for i from 1 to 2 do
 for j from 1 to 2 do
  if A2[i,j] mod (1/2) = True then
   counterd := counterd + 1:
  end if:
 od:
od:
counterd;
if counterd = 4 then
 for i from 1 to 2 do
 for j from 1 to 2 do
  A2[i,j] := A2[i,j]/(1/2):
 od:
od:
end if:
Error, invalid input: modp received 1/2, which is not valid for its 2nd argument, m;
 
i do it in another method
mod has error in maple 12
 
how to mod in maple 12 for fraction ?

@Kitonum 

moreover , is it possible sqrt(3)/6 and 1/2 combine into one term and the rest is one matrix result?

@Kitonum 

Your doing is correct.

A := Matrix(2, 2, {(1, 1) = (1/6)*sqrt(3)+(1/2)*I, (1, 2) = (1/6)*sqrt(3)-(1/2)*I, (2, 1) = (1/6)*sqrt(3)-(1/2)*I, (2, 2) = (1/6)*sqrt(3)+(1/2)*I}):
A1:=(sqrt(3)/6)%*(Matrix(2,(i,j)->Re(A[i,j]))/(sqrt(3)/6))%+((1/2)%*(Matrix(2,(i,j)->A[i,j]-Re(A[i,j]))/(1/2)));
 
Error, (in rtable/Product) invalid arguments
in maple 12
 
how to run in maple 12?
 
 

@ThU 

your doing is not factorize, or collect

i mean apply on this example matrix

Matrix(2, 2, {(1, 1) = (1/6)*sqrt(3)+(1/2)*I, (1, 2) = (1/6)*sqrt(3)-(1/2)*I, (2, 1) = (1/6)*sqrt(3)-(1/2)*I, (2, 2) = (1/6)*sqrt(3)+(1/2)*I})

expect a general method.

and the expected output is like

Matrix([[1,-I],[-I,1]])

though this is wrong matrix for above example

actually i am finding like pauli matrix

hope to extract (1/6)*sqrt(3),  (1/2) out of matrix and leave matrix like

Matrix([[1,-I],[-I,1]]), one or zero or imaginary 1

@vv 

i have already finished 

and would like to delete this question

but it do not allow to delete question as there are reply

what should I do ?

@vv 

in previous post, there is definition

i searched that there are extra 7 possible matrices 

need further research whether these 7 matrices are useful or not.

@Joe Riel 

i change to use printf to print matrix

@vv 

in fact, I find extra 7 matrix, possible are new hope besides 1 i has known from observation.

@vv 

i successfully found invariant matrix 

besides zero matrix, there are also one and zero matrix

but it is just the beginning, i am very excited.

you want to know more what i will do?

@vv 

i successfully find using another method.

 

@vv 

final := choose(result, 10):

result in above is to store basic matrices after elementary operations

i try to make all different kind of elementary operation on matrix

then choose 10 equations,

why 10 equation?

for example if 3 equations , and choose 1 at right hand side of equal, eq1 = eq1 will make 3 equations system to 2 equations system

[eq1,eq2,eq3]

[eq2 = eq1, eq3=eq1, eq1=eq1]

so, to make 9 equations for 9 variables, need to choose 10 equations,

however using double for loop, it will result a list with number of equations more than 9,

so choose again for 9 different equations only.

@vv 

looking for Invariant matrix

this wiki is correct.

@acer 

there is a case that we may not find

if power of 2 term has coefficient

such as 5*2^3+7*2^5 etc

is it possible to find?

if further assume these coefficients are prime only, can this be find?

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