Fundament
In MPkn: Calculations of One Discrete Model in Several Time Steps

Basis (Fundament) in the Package of the Methods Used

Basic: Step in the Chain of Power of Two

$M_1 = A$
$M_2 = (I + Q)M_1 = A + QA$
$M_3 = A + QA + Q^2A$
$M_4 = A + QA + Q^2A + Q^3A = M_2 + (Q^2 + Q^3)A$
$M_4 = M_2 +Q^2(I + Q)A = M_2 + Q^2M_2$
$M_4 = (I + Q^2)M_2$
$M_5 = M_4 + Q^4A$
$M_6 = M_5 + Q^5A$
$M_7 = M_6 + Q^6A$
$M_8 = M_7 + Q^7A = M_6 + (Q^6 + Q^7)A$
$M_8 = M_5 + (Q^5 + Q^6 + Q^7)A = M_4 + (Q^4 + Q^5 + Q^6 + Q^7)A$
$M_8 = M_4 + Q^4(I + Q + Q^2 + Q^3)A$
$M_8 = M_4 + Q^4M_4 = (I + Q^4)A$
$...$
$M_{2n} = (I + Q^n)M_n, n = 1, 2, 3, ..$

More Complex

$M_3 = A + QA + Q^2A$
$M_4 = A + QA + Q^2A + Q^3A$
$M_5 = A + QA + Q^2A + Q^3A + Q^4A$
$M_6 = A + QA + Q^2A + Q^3A + Q^4A + Q^5A$
$M_6 = (I + Q^3)M_3$
$...$
$M_{32^n} = (I + Q^{32^{n-1}})M_{3*2^{n-1}}$,
$n = 1, 2, 3, ..$

and

$M_{k{2^{n}}} = (I + Q^{k2^{n-1}})*M_{k2^{n-1}}$,
$k = 1, 2, 3, .., n = 1, 2, 3, ..$

Examples

For Basic

require(matrixcalc)
An = 2
matmult <- function(A, B){
    C = matrix(numeric(4), 2, 2)
    for (i in 1:2){
        for (j in 1:2){ C[i, j] = sum(A[i, ]*B[, j])}
    }
    return(C)
}
Q = array(c(0.58, 0.53, 0.42, 0.47), c(2, 2))
q = 0
for (i in 1:8){
  q = q + matrix.power(Q, i)
}
print(paste("i =", i))
print(q)
M = Q
I = diag(1, 2, 2)
n = c(1, 2, 4, 8)
for (i in 2:length(n)){
  M = matmult((I + matrix.power(Q, n[i-1])), M)
}
print(paste("n[i] =", n[i]))
print(M)

For More

```r
Q = array(c(0.58, 0.53, 0.42, 0.47), c(2, 2)) k = 3 q = 0 for (i in 1:48){ q = q + matrix.power(Q, i) if (i == k) Qk = q } print(paste("i =", i)) print(q)

M = Qk I = diag(1, 2, 2) n = integer(5) for (i in 1:5){ n[i] = k*2^(i-1) } print(n) for (i in 2:length(n)){ M = matmult((I + matrix.power(Q, n[i-1])), M) } print(paste("n[i] =", n[i])) print(M)