We assume that it is possible to calculate finite numerical integration via statistical matrix mechanics (a product of the statistical method of Cairo techniques).
It is obvious that here it is not necessary to use FDM or any similar calculation method [1].
The finite statistical integration for 1D,2D,3D space is not complicated but a little long.
The starting point is to calculate the transfer matrix Dnxn given by the following expression [1],
Dnxn=1/(I-Bnxn)-I
Where B is the well-defined statistical transition matrix.
Then calculate the numerical statistical integration weights (Wi, j) of the bounded area (G) of x, y as follows:
Wi,j=Σ (sigma) D(i, j) on j/(Σ Σ D(i,j) on i, j.
Note that i, j replaces x, y.
Now, the statistical integration Ist of any function f(x, y) over the finite domain G will be given by,
Ist=Σf(x, y).(Wi, j) for all i, j.
It is worth mentioning that the first statistical integration Ist, although not complex, is more accurate than any other FDM technique.
[1]Effective unconventional approach to statistical differentiation and statistical integration, ResearchGate, IJISRT journal,
November 2022.
Hello, check this out, it addresses the finite element method using matrices (Chapters 15 and 18 essentially).
https://www.researchgate.net/publication/377566235_Applied_Mathematics_Theories_Methods_and_Practices_From_simple_to_Complex_Problems_For_Beginners_and_Experts
This free paper could also help: https://www.iaeng.org/publication/WCE2015/WCE2015_pp105-110.pdf
This is a brief answer to shed some light on the question and its answer.
A-Consider a one-dimensional finite integer I=
∫ Y dx from x=a to x=b where Y is an arbitrary function of x defined on [a,b] and the domain [a,b] is divided into n=7 equidistant free nodes.
Note that n is chosen arbitrarily and can be odd or even and that the precision increases as n increases.
Here the well defined tri-diagonal transition matrix B is given by the expression,
Bnxn =
0 .5 0 0 0 0 0
.5 0 .5 0 0 0 0
0 .5 0 .5 0 0 0
0 0 .5 0 .5 0 0
0 0 0 .5 0 .5 0
0 0 0 0 .5 0 .5
0 0 0 0 0 .5 0
Then I-B matrix will be given by,
1 -1/2 0 0 0 0 0
-1/2 1 -1/2 0 0 0 0
0 -1/2 1 -1/2 0 0 0
0 0 -1/2 1 -1/2 0 0
0 0 0 -1/2 1 -1/2 0
0 0 0 0 -1/2 1 -1/2
0 0 0 0 0 -1/2 1
and the transfer matrix Dnxn = [1/(I-B)]-I is expressed as follows,
3/4 3/2 5/4 1 3/4 1/2 1/4
3/2 2 5/2 2 3/2 1 1/2
5/4 5/2 11/4 3 9/4 3/2 3/4
1 2 3 3 3 2 1
3/4 3/2 9/4 3 11/4 5/2 5/4
1/2 1 3/2 2 5/2 2 3/2
1/4 1/2 3/4 1 5/4 3/2 3/4
In fact the transfer matrix Dnxn is the statistical integration matrix.
Here we have, Σ (sigma) D(i, j) on j is calculated as,
(6,11,14,15,14,11,6)
for i=1 to 7 respectively.
And the double summation Σ Σ D(i,j) on i, j is calculated as follows:
(6+11+14+15+14+11+6) =77
Therefore, the seven statistical weights for 7 equidistant free nodes are given by:
{6/77.11/77.14/77.15/77.14/77.11/77.6/77}
and the statistical integration formula is expressed as follows:
Ist = [6/77*Y1+11/77*Y2+14/77*Y3+15/77*Y4 +14/77*Y5+11/77*Y6+6/77*Y7]
To be continued.
Answer II – continued
B-Consider a finite two-dimensional integer I on the finite two-dimensional domain G
I=∫∫ Z dx dy from x =a to x=b and from y=c to y=d.
Where G is the rectangular two-dimensional domain of x∈ [a,b],y ∈ [c,d] and Z is an arbitrary function of x,y defined on the domain G.
The domain G is discretized into nine equidistant free nodes.
We follow exactly the same procedure described in section A to obtain the transfer matrix Dnxn = D9x9 as that of case A.
The numerical results for the I-B matrix are,
1 -1/4 0 -1/4 0 0 0 0 0
-1/4 1 -1/4 0 -1/4 0 0 0 0
0 -1/4 1 0 0 -1/4 0 0 0
-1/4 0 0 1 -1/4 0 -1/4 0 0
0 -1/4 0 -1/4 1 -1/4 0 -1/4 0
0 0 -1/4 0 -1/4 1 0 0 -1/4
0 0 0 -1/4 0 0 1 -1/4 0
0 0 0 0 -1/4 0 -1/4 1 -1/4
0 0 0 0 0 -1/4 0 -1/4 1
Then the transfer matrix D9x9 = [1/(I-B) -I] is calculated and the results are as follows:
11/56 11/28 1/8 11/28 1/4 3/28 1/8 3/28 3/56
11/28 9/28 11/28 1/4 1/2 1/4 3/28 5/28 3/28
1/8 11/28 11/56 3/28 1/4 11/28 3/56 3/28 1/8
11/28 1/4 3/28 9/28 1/2 5/28 11/28 1/4 3/28
1/4 1/2 1/4 1/2 1/2 1/2 1/4 1/2 1/4
3/28 1/4 11/28 5/28 1/2 9/28 3/28 1/4 11/28
1/8 3/28 3/56 11/28 1/4 3/28 11/56 11/28 1/8
3/28 5/28 3/28 1/4 1/2 1/4 11/28 9/28 11/28
3/56 3/28 1/8 3/28 1/4 11/28 1/8 11/28 11/56
In fact the D9x9 transfer matrix is the statistical integration matrix.
Here we have, Σ (sigma) D(i, j) on j is calculated as,
(98,140,98,140,196,140,98,140,98)/56
for i=1 to 9+9
respectively.
And the double summation Σ Σ D(i,j) on i, j is calculated as follows:
(98+140+98+140+196+140+98+140+98)/56=20.5
Therefore, the nine statistical weights for 9 equidistant free nodes in two-dimensional geometry are given by:
{0.0854,0.122,0.0854,0.122,0.171,0.0854,0.1222,0.0854}
and the statistical integration formula is expressed as follows:
Ist = [0.0854*Z(1,1)+0.122*Z(2,1)+0.0854*Z(3,1)+0.122*Z(2,1)+0.171*Z(2,2 )+0.0854*Z(2,3)+0.0854*Z(3,1)+0.122*Z(3,2)+0.0854*Z(3,3)] . 9h^2
Where h is the interval between two adjacent nodes.
To be continued.
Science leaves the era of mathematics and enters the era of matrix mechanics and the turning point is the discovery of numerical statistical theory called Cairo techniques and its transition matrices eligible to solve almost all problems of classical physics and of quantum mechanics.
No more partial differential equations, no more numerical integration, no more FDM techniques. . etc.
Statistical matrix mechanics is capable of solving all of the above problems and additionally predicting unknown universal physical rules.
Note that the Heisenberg matrix and the Dirac matrix are neither physical nor statistical and incomplete.
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