Funny pictures of sexps

[bo198214]

Actually, I wanted to confirm something, because it's been a while since I last tried my hand at this issue (recentering).

We have the Abel function, A(x) = A(f(x))+1, correct? Here, f(x) is e^x. To solve, we find A(f(x))-A(x) = 1, and this in turn implies A*(C - I) = 1, where C is the Carleman matrix of f(x), I is the identity matrix, and A is the power series coefficients of the Abel function A, and 1 is the vector [1, 0, ...].

D'acord.

So to recenter, we have to substitute u(x) = x+1, for example. This implies that we solve A(f(u(x))) - A(u(x)) = 1, so A(Cu - Iu) = 1.

So Cu is the the Carleman matrix of e^(x+1), and Iu will be a suitable triangular Pascal matrix, the Carleman matrix of u(x).

My recentering takes place at the function level not on the (truncated) Matrix level.
We have a method to compute the Abel function of f that uses the powerseries expansion of f at 0.

Now we want to apply this method to a different development point x0.
Then I first move the development to 0.
g(x)=f(x+x0)-x0
or if \( \tau(x)=x+x0 \) we can also write it as:
\( g = \tau^{-1} \circ f \circ \tau \)
Now I can apply the method and find an Abel function \( \alpha \) of g.
It satisfies:
\( \alpha\circ g = s\circ \alpha \) with \( s(x)=x+1 \).
Thatswhy
\( \alpha\circ \tau^{-1} \circ \underbrace{\tau \circ g \circ \tau^{-1}}_{=f} = s\circ \alpha \circ\tau^{-1} \)

And so \( \alpha\circ \tau^{-1} \) or \( x\mapsto \alpha(x-x0) \) is an Abel function of \( f \).

This is completely independent on matrices. And one can theoretically do it also with other functions \( \tau \) than translations.