By the way, I had another observation to add.

For the slog, the singularities at the primary fixed points look like ordinary logarithms. I'm still trying to figure out the best way to state this. My latest addition is to make explicit that this holds in a disk around the fixed point with a radius greater than zero but tending towards zero.

Anyway, if we "remove" these singularities, the root test seems to be tending to 0.72, indicating that the "residue" as I have called it still has a singularity of some sort, most likely at the fixed points. I say most likely, but I know that's where the singularities are, just by casual analysis of the slog. You see, the branches aren't like the branches of the ordinary logarithm. In the ordinary logarithm, each point in two branches differ by the same constant. Stated more simply, the first and all subsequent derivatives are equal at the same point in different branches.

However, with the slog, this is not true. The first few derivatives are different, and I'm willing to wager that all derivatives are different. Hence, even after subtracting out the ordinary logarithm at the fixed point, we still have branches and the equivalent of branch cuts as seen from any point of view. However, rather than differing by constant amounts, the points on neighboring branches differ by an amount that is 0 at the fixed point and increasing as we move away from the fixed point. There is precedent for this: consider , which has two branches (it's cyclic as we go around the origin). At 0.01, the values are +0.001 and -0.001, and at 1, the values are +1 and -1.

The point is, while the slog looks like an ordinary logarithm at either fixed point, we can't eliminate the branch cuts by subtracting the logarithm, and hence, we can't reduce the root test (thereby increasing the radius of convergence).

However, I think we can subtract the logarithm at sexp(-2)! If we consider a loop around z=-2, with a small radius, then we can look at the graph of the slog to see what will happen. (By a "small radius", I mean less than 1 for sure, though something between 0.5 and 0.75 makes it easy to see on the slog graph.)

The slog of this loop with fixed radius will look like a slightly wavy vertical line, and it allows us to move from branch to branch of the slog as we go up or down multiples of 2*pi*i.

Each major branch of the slog (those right off the "backbone") looks exactly like any other major branch (due to cyclic symmetry), and hence, the branches of the sexp will look the "same" as we loop the singularity at z=-2. Therefore, subtracting the natural logarithm at z=-2 should eliminate the branch cuts due to this singularity. Therefore, the radius of convergence of the slog with the removed natural logarithm should be limited by the singularity at z=-3. Accordingly, the root test of the power series for sexp(z-1) should be 1, but the root test for sexp(z-1)-ln(z+1) should be 0.5, indicating a radius of convergence of 2.

Furthermore, with a bit of effort, we should be able to determine the coefficients for an ideal version of the singularity at z=-3, allowing us to subtract them out, and get an even smaller "residue" of the sexp. I've already confirmed that the singularity at z=-3 (effectively, something along the lines of ln(ln(z-3)) or so) has non-matching branches, so even after removing this singularity, we'll most likely still have a branch cut and hence a limited radius of convergence. But we should be able to get fairly accurate coefficients, at least beyond the first dozen or so, and perhaps these could be fed into an iterative solver for the non-linear system solution of the sexp. Or perhaps, after going that far, further insights will be within reach...

For the slog, the singularities at the primary fixed points look like ordinary logarithms. I'm still trying to figure out the best way to state this. My latest addition is to make explicit that this holds in a disk around the fixed point with a radius greater than zero but tending towards zero.

Anyway, if we "remove" these singularities, the root test seems to be tending to 0.72, indicating that the "residue" as I have called it still has a singularity of some sort, most likely at the fixed points. I say most likely, but I know that's where the singularities are, just by casual analysis of the slog. You see, the branches aren't like the branches of the ordinary logarithm. In the ordinary logarithm, each point in two branches differ by the same constant. Stated more simply, the first and all subsequent derivatives are equal at the same point in different branches.

However, with the slog, this is not true. The first few derivatives are different, and I'm willing to wager that all derivatives are different. Hence, even after subtracting out the ordinary logarithm at the fixed point, we still have branches and the equivalent of branch cuts as seen from any point of view. However, rather than differing by constant amounts, the points on neighboring branches differ by an amount that is 0 at the fixed point and increasing as we move away from the fixed point. There is precedent for this: consider , which has two branches (it's cyclic as we go around the origin). At 0.01, the values are +0.001 and -0.001, and at 1, the values are +1 and -1.

The point is, while the slog looks like an ordinary logarithm at either fixed point, we can't eliminate the branch cuts by subtracting the logarithm, and hence, we can't reduce the root test (thereby increasing the radius of convergence).

However, I think we can subtract the logarithm at sexp(-2)! If we consider a loop around z=-2, with a small radius, then we can look at the graph of the slog to see what will happen. (By a "small radius", I mean less than 1 for sure, though something between 0.5 and 0.75 makes it easy to see on the slog graph.)

The slog of this loop with fixed radius will look like a slightly wavy vertical line, and it allows us to move from branch to branch of the slog as we go up or down multiples of 2*pi*i.

Each major branch of the slog (those right off the "backbone") looks exactly like any other major branch (due to cyclic symmetry), and hence, the branches of the sexp will look the "same" as we loop the singularity at z=-2. Therefore, subtracting the natural logarithm at z=-2 should eliminate the branch cuts due to this singularity. Therefore, the radius of convergence of the slog with the removed natural logarithm should be limited by the singularity at z=-3. Accordingly, the root test of the power series for sexp(z-1) should be 1, but the root test for sexp(z-1)-ln(z+1) should be 0.5, indicating a radius of convergence of 2.

Furthermore, with a bit of effort, we should be able to determine the coefficients for an ideal version of the singularity at z=-3, allowing us to subtract them out, and get an even smaller "residue" of the sexp. I've already confirmed that the singularity at z=-3 (effectively, something along the lines of ln(ln(z-3)) or so) has non-matching branches, so even after removing this singularity, we'll most likely still have a branch cut and hence a limited radius of convergence. But we should be able to get fairly accurate coefficients, at least beyond the first dozen or so, and perhaps these could be fed into an iterative solver for the non-linear system solution of the sexp. Or perhaps, after going that far, further insights will be within reach...

~ Jay Daniel Fox