Understanding Abel/Schroeder with matrix-expression

[Gottfried]

I am not 100% sure what you are meaning, but I think you convey the notion of \( b^{\dots ^{b^c}} \) to \( u*\dots *u * p_b( c) \).
But thats probably not adequate.
When you do regular iteration the ordering is already contained.
It is about \( h \) time applying a function to an argument.
If the function is exponentiation (letting aside the fixed point shift) then this just means that the argument is on top of the tower and nowhere else.
The method only works for this case.
There is no freedom to rearrange the tower, say be rearring the \( u \)'s and \( p_b( c) \).

Was it that?

Well, first I've to say, that possibly I'm completely off the road with my "brainstorm", since I've still far too little experience with the concept of the Schroeder-functional equation.

I refer here to your article there

For doing this we first compute the regular Schroeder function (note that the Schroeder function is determined up to a multiplicative constant and the Abel function is determined up to an additive constant). A Schroeder function \( \sigma \) of a function \( f \) is a function that satisfies the Schroeder equation

\( \sigma(f(x))=s\sigma(x) \)

(...)
This function particularly yields the regular iteration at 0, via \( f^{\circ t}(x)=\sigma^{-1}(s^t\sigma(x)) \).
(...)

and my notation of \( p_b(x) \) is meant to be \( \sigma(x) \), of \( q_b(x) \) to be \( \sigma^{-1}(x) \) and of \( u^h \) to be \( s^t \) .

Here b, as usual, is the base, \( t^{1/t}=b \), u=log(t) and I assume my formula-substitution to get the Schroeder-equation \( f_b^{^oh}(x)= q_b( u^h*p_b(x)) =\sigma^{-1}(s^h\sigma(x)) \) from the diagonalization-method.


If I assume the transposed carleman-matrix Bb for x->b^x as diagonalizable, then I also assume formally for any iteration-height h

\( \hspace{24} B_b^h = W_b * dV(u)^h * W_b^{-1} \)

where Wb contains the set of eigenvectors of Bb.

Then the whole operation

\( \hspace{24} V(x)\sim * B_b^h = V(\exp_b^{^{o}h}(x))\sim \)

can be splitted into

\( \hspace{24} V(x)\sim * W_b * dV(u)^h * W_b^{-1} = V(\exp_b^{^oh}(x))\sim \)

Here it would be nice, if Wb behaves as an "operator on powerseries", such that

\( \hspace{24} V(x)\sim * W_b = V(y)\sim \)

If this would be so, then also Wb^-1 would be such an operator and all argumentation simplifies.


Empirically for some convergent cases this seems to be true, so assuming this we can split our computation into three parts:
\( \hspace{24} V(x)\sim * W_b = V(y)\sim \)
\( \hspace{24} V(y)\sim * dV(u)^h = V(z)\sim \)
\( \hspace{24} V(z)\sim * W_b^{-1} = V(\exp_b^{^oh}(x))\sim \)

The value for y can be seen as evaluation of a powerseries in x, which uses the coefficients of the second column of Wb, and I denoted this as function \( y=p_b(x) \), which I understood as the function \( \sigma(x)) \) in your text.

Then \( V(y)\sim *dV(u)^h \) is \( V(p_b(x)*u^h)~ \) because both V(y) and V(u)^h are vandermonde-vectors and dV(u)^h is diagonal.
Then since Wb^-1 provides the coefficients for the inverse \( q_b(x) = p_b^{-1}(x) \) we have finally

\( \hspace{24} V(p_b(x)*u^h)~ *W_b^{-1} = V(q_b(u^h*p_b(x)))~ \)

which looks like a perfect match with the Schroeder-formula

\( \hspace{24} = V(\sigma^{-1}(s^h * \sigma(x)))~ \)

---------------------------------

So, what is the relation to my previous post?

I confess, I've difficulties myself to restate the problem as I've seen it in the morning. But let's try.

I was considering a sequence of computations

\( \hspace{24} f_b^{{^o}1}(x)= \sigma^{-1}(s*\sigma(x)) \)
\( \hspace{24} f_b^{{^o}2}(x)= \sigma^{-1}(s*\sigma(\sigma^{-1}(s*\sigma(x)))) = \sigma^{-1}(s*s*\sigma(x)) \)
...

and asked, why the heck does \( \sigma^{-1} \) know, that \( \sigma(x) \) has to appear at the top since all factors of this scalar equation have the same level and can arbitrarily be interchanged in the above sequence of computations...

Hmm...