10/27/2007, 06:32 AM
jaydfox Wrote:Then just today, I was thinking about the sexp function, having finally decided to turn my attention back to it a couple days ago. Anyway, I was thinking about the singularity at sexp(-2). It occurred to me that in the immediate vicinity of -2, it would look pretty much like \( \ln(a_1 z) \), where \( a_1 \) is the coefficient for the first degree term in the power series for sexp. In fact, this would just reduce to \( \ln(z)+\ln(a_1) \).I suppose this isn't so interesting if you take the power series for sexp(z-1). After all, if you assumed a generic linear critical interval (-1, 0), or even a simple third order approximation, then the interval (-2, -1) would pretty much look like a logarithm, and as such, the power series at -1 (from the left) would pretty much be that of a logarithm.
As such, near the singularity, and assuming no other singularities in the immediate vicinity, the power series for sexp should approximately equal the power series for the natural logarithm.
Sure enough, I took my terms for the power series of the slog at 0, and calculated the reversion of the series to get the power series for the sexp (at -1). When I calculated the power series for the first derivative (a trivial calculation), I found that after the first half dozen terms, the terms of the first derivative of the sexp are alternating plus or minus 1 to within 1% or less, and by the 18th term, they're equal to +/- 1 to within 1 part in a million.
In other words, the terms of the power series of sexp(z-1) converge on the terms for the power series of ln(z+1).
But where it would get interesting is when you take power series further and further to the right. The power series at z=0, z=1, z=2, etc., would start out looking more and more like iterated exponentials, yet they would still converge on the power series of a logarithm with its singularity at -2.
~ Jay Daniel Fox

