A nice accidental result - although it is nearly trivial...
I tried, which matrix-operator would perform the left-associative exponentiation - well, this can easily be solved, but the appearance of this operator was much surprising to me.
First assume the factorially similarity-scaled matrices of Stirling numbers 2nd and 1st kind (as I described them many times). If S2 is that lower-triangle containing Stirling-numbers 2nd kind and S1 that containing 1st kind, then fS2F and fS1F are the factorially similarity scalings:
fS1F = dF^-1 * S1 * dF
fS2F = dF^-1 * S2 * dF
for instance:
Now consider the matrix
If we use it as matrix-operator we have the following transform of powerseries:
so
and integer iterations by integer powers of DPW_m
If we interpret V(x-1) as a binomial-transform of V(x) then we can note this at the beginning and at end we have:
and denote the new product of matrices in the parenthese formally
then we have the operator for x-> x^m or, when k-times iterated, x -> x^(m^k), whenever the product P^-1~ * () is computable.
The much surprising aspect is the shape of DPW_m * P~: this is just the binomial-matrix where the columns represent the binomial-coefficients for the orders according to m - which also can be fractional.
For instance, m=2, DPW_2*P~ =
which gives, premultiplied by P^-1~ (here the multiplication is possible) the "quasi"-diagonal matrix, which simply shifts all powers of x in V(x) to their 2*power-position V(x^2).
For instance, the even-indexed powers in [(1),x,(x^2),x^3,(x^4)...] get simply shifted to the consecutive positions [(1),(x^2),(x^4),(x^6),...], so we have a transform of x->x^2 and of all consecutive powers.
For fractional "bases of iteration" ( this means exponents here) m=1/2
the premultiplication by P^-1~ gives divergent results, cannot be done and must applied to the V(x)~-parameter instead.
The most interesting aspect is, that with integer m the column-vectors of PW_m are finite - so we can safely define the version with P^-1~ premultiplied - which gives, for integer m (and heights) - nicely the quasi-diagonal-matrices, which transform a vandermondevector V(x) into one of V(x^2), V(x^3) or the like...
And finally, the aspect, which really pleases me, is, that for fractional m we just get the correct column-vectors of fractional binomials - a smooth procedure, which I was looking for over the last monthes (unsatisfied with the hardcoded binomial-expressions otherwise)
Here another example with m=2/3, which means, each multiple of third power of x in the result is composed by smaller integer powers of x, and only the other powers are composed by the infinite sequences of according fractional binomially weighted powers of x:
so, the spin-off-result here is:
the fractional m in fS1F*dV(m)*fS2F*P~ gives the correct composition for the matrix-operator, which performs the binomial-theorem for fractional exponents, and its (possibly fractional) powers give the expected formal powerseries for the expected binomial composition.
I tried, which matrix-operator would perform the left-associative exponentiation - well, this can easily be solved, but the appearance of this operator was much surprising to me.
First assume the factorially similarity-scaled matrices of Stirling numbers 2nd and 1st kind (as I described them many times). If S2 is that lower-triangle containing Stirling-numbers 2nd kind and S1 that containing 1st kind, then fS2F and fS1F are the factorially similarity scalings:
fS1F = dF^-1 * S1 * dF
fS2F = dF^-1 * S2 * dF
for instance:
Code:
´
fS2F = fS1F^-1
V(x)~ * fS1F = V(log(1+x))~
Now consider the matrix
Code:
´
DPW_m = fS1F * dV(m) * fS2F
If we use it as matrix-operator we have the following transform of powerseries:
Code:
´
V(x-1)~ * fS1F = V(log(x))~
V(log(x))~ * dV(m) = V(m*log(x))~ = V(log(x^m))~
V(log(x^m))~ * fS2F = V(exp(log(x^m))-1) ~
= V(x^m -1)~
Code:
´
V(x - 1)~ * DPW_m = V(x^m - 1)~
and integer iterations by integer powers of DPW_m
Code:
´
V(x - 1)~ * DPW_m^k = V(x^(m^k) - 1)~
If we interpret V(x-1) as a binomial-transform of V(x) then we can note this at the beginning and at end we have:
Code:
´
(V(x)~ * P^-1~) * DPW_m^k = (V(x^(m^k))~ * P^-1 ~)
then
V(x)~ * (P^-1~ * DPW_m^k * P~) = V(x^(m^k))~
and denote the new product of matrices in the parenthese formally
Code:
´
PW_m = P^-1 ~ * (DPW_m * P~)
then we have the operator for x-> x^m or, when k-times iterated, x -> x^(m^k), whenever the product P^-1~ * () is computable.
The much surprising aspect is the shape of DPW_m * P~: this is just the binomial-matrix where the columns represent the binomial-coefficients for the orders according to m - which also can be fractional.
For instance, m=2, DPW_2*P~ =
Code:
´
1 1 1 1 1 1 1 1
. 2 4 6 8 10 12 14
. 1 6 15 28 45 66 91
. . 4 20 56 120 220 364
. . 1 15 70 210 495 1001
. . . 6 56 252 792 2002
. . . 1 28 210 924 3003
. . . . 8 120 792 3432
which gives, premultiplied by P^-1~ (here the multiplication is possible)
Code:
´
1 . . . . . . .
. . . . . . . .
. 1 . . . . . .
. . . . . . . .
. . 1 . . . . .
. . . . . . . .
. . . 1 . . . .
. . . . . . . .
For instance, the even-indexed powers in [(1),x,(x^2),x^3,(x^4)...] get simply shifted to the consecutive positions [(1),(x^2),(x^4),(x^6),...], so we have a transform of x->x^2 and of all consecutive powers.
For fractional "bases of iteration" ( this means exponents here) m=1/2
Code:
´
1 1 1 1 1 1 1 1
. 1/2 1 3/2 2 5/2 3 7/2
. -1/8 . 3/8 1 15/8 3 35/8
. 1/16 . -1/16 . 5/16 1 35/16
. -5/128 . 3/128 . -5/128 . 35/128
. 7/256 . -3/256 . 3/256 . -7/256
. -21/1024 . 7/1024 . -5/1024 . 7/1024
. 33/2048 . -9/2048 . 5/2048 . -5/2048
The most interesting aspect is, that with integer m the column-vectors of PW_m are finite - so we can safely define the version with P^-1~ premultiplied - which gives, for integer m (and heights) - nicely the quasi-diagonal-matrices, which transform a vandermondevector V(x) into one of V(x^2), V(x^3) or the like...
And finally, the aspect, which really pleases me, is, that for fractional m we just get the correct column-vectors of fractional binomials - a smooth procedure, which I was looking for over the last monthes (unsatisfied with the hardcoded binomial-expressions otherwise)
Here another example with m=2/3, which means, each multiple of third power of x in the result is composed by smaller integer powers of x, and only the other powers are composed by the infinite sequences of according fractional binomially weighted powers of x:
Code:
´
1 1 1 1 1 1 1 1
. 2/3 4/3 2 8/3 10/3 4 14/3
. -1/9 2/9 1 20/9 35/9 6 77/9
. 4/81 -4/81 . 40/81 140/81 4 616/81
. -7/243 5/243 . -10/243 35/243 1 770/243
. 14/729 -8/729 . 8/729 -14/729 . 308/729
. -91/6561 44/6561 . -28/6561 35/6561 . -154/6561
. 208/19683 -88/19683 . 40/19683 -40/19683 . 88/19683
so, the spin-off-result here is:
the fractional m in fS1F*dV(m)*fS2F*P~ gives the correct composition for the matrix-operator, which performs the binomial-theorem for fractional exponents, and its (possibly fractional) powers give the expected formal powerseries for the expected binomial composition.
Gottfried Helms, Kassel