Funny pictures of sexps

[jaydfox]

Bo, just to refresh my memory, how are you calculating the isexp_e?

I compute the powerseries of Andrew's slog at x=1. Then I invert this series.
To be more exact:
I consider the Carleman matrix C of the powerseries development of e^(x+1)-1.
Then I compute what Andrew calls Abel matrix.
I.e. the transpose of the suitably truncated C - I. And solve the equation A p = (1,0,0...)
-1+p(x) is then the islog of e^(x+1)-1 at 0.
And p(x-1) is the islog of e^x at 1.
And p^{-1}(x)+1 is the isexp of e^x at 1.

The development of the slog at 1 instead of the normal 0, which implies sexp developed at 0 instead of the normal -1, should allow a convergence radius of 2 instead of 1 for the sexp. But it seems as if the convergence radius is only 1.4, i.e. roughly the imaginary part of the fixed point.

edit: On the other hand probably the at 1 developed slog is different from the at 0 development slog.

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, ...].

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).

This does not seem to match your description of the steps you took, so I wanted to be sure that I wasn't missing something, or that I hadn't misunderstood your steps. I fully acknowledge that I could have worked this out incorrectly.