I finally bothered to put in the effort to learn how to use power series in SAGE. The native reversion function says it's not implemented, so I ended up learning how to use PARI's power series implementation. There was a small learning curve, but I think I'm getting the hang of it.

As it turns out, for f(x) = ln(ln(x)), the singularity at x=1 is just a plain natural logarithm, and it can be removed. To see this, construct a power series for ln(ln(x+2))-ln(x+1). If the singularity at x=1 were not fully removed (if branch cuts remained), then the root test would converge on 1, even if it were initially a bit smaller.

However, the root test is 0.5, indicating a radius of convergence of 2. The singularity at x=1 is completely removed, even if technically the function is not analytic at that point.

Therefore, we should be able to subtract ln(ln(z+2)) from sexp(z-1), and get a "residue" (I know I need a better name for it). This residue will have a radius of convergence of 2, just like sexp(z-1)-ln(z+1), but nonetheless the terms are a couple orders of magnitude more accurate.

In fact, I've taken the liberty of making a chart:

On this chart, I've calculated the logarithm, base 2, of each term in the power series. I chose base 2, because a slope of -1 indicates a root test of 1/2, while a slope of 0 indicates a root test of 1.

As you can see, the red graph is essentially flat, giving a root test of 1, due to the singularity at z=-2. The blue and green graphs both converge on a slope of -1, so they both have a root test of 0.5, and a radius of convergence of 2. So the second singularity wasn't completely removed. But the green graph is noticeably lower than the blue. In fact, by the 400th term in the series, the green chart is about 6.21 units below the blue chart, indicating that the green coefficients are almost 75 times smaller (2^6.21). Therfore, the green coefficients represent a better "residue".

The difference is small enough, however, that if calculation speed is an issue, the blue coefficients can be used.

For example, given these three sets of coefficients, there are three ways to calculate the sexp (though it must be within a radius of 2, centered at z=-1, and a radius of 1 for the first equation):

For the greatest accuracy, use the third equation. For greater speed, with little loss of precision, use the second equation. The first equation is mainly just for reference, and should be avoided, unless you're calculating values very close to z=-1.

And if you're wanting to attempt analytic continuation, the definitely use the second or third equation. The larger radius of convergence means you can move twice as far for the same loss of series precision.

jaydfox Wrote: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...

As it turns out, for f(x) = ln(ln(x)), the singularity at x=1 is just a plain natural logarithm, and it can be removed. To see this, construct a power series for ln(ln(x+2))-ln(x+1). If the singularity at x=1 were not fully removed (if branch cuts remained), then the root test would converge on 1, even if it were initially a bit smaller.

However, the root test is 0.5, indicating a radius of convergence of 2. The singularity at x=1 is completely removed, even if technically the function is not analytic at that point.

Therefore, we should be able to subtract ln(ln(z+2)) from sexp(z-1), and get a "residue" (I know I need a better name for it). This residue will have a radius of convergence of 2, just like sexp(z-1)-ln(z+1), but nonetheless the terms are a couple orders of magnitude more accurate.

In fact, I've taken the liberty of making a chart:

On this chart, I've calculated the logarithm, base 2, of each term in the power series. I chose base 2, because a slope of -1 indicates a root test of 1/2, while a slope of 0 indicates a root test of 1.

As you can see, the red graph is essentially flat, giving a root test of 1, due to the singularity at z=-2. The blue and green graphs both converge on a slope of -1, so they both have a root test of 0.5, and a radius of convergence of 2. So the second singularity wasn't completely removed. But the green graph is noticeably lower than the blue. In fact, by the 400th term in the series, the green chart is about 6.21 units below the blue chart, indicating that the green coefficients are almost 75 times smaller (2^6.21). Therfore, the green coefficients represent a better "residue".

The difference is small enough, however, that if calculation speed is an issue, the blue coefficients can be used.

For example, given these three sets of coefficients, there are three ways to calculate the sexp (though it must be within a radius of 2, centered at z=-1, and a radius of 1 for the first equation):

For the greatest accuracy, use the third equation. For greater speed, with little loss of precision, use the second equation. The first equation is mainly just for reference, and should be avoided, unless you're calculating values very close to z=-1.

And if you're wanting to attempt analytic continuation, the definitely use the second or third equation. The larger radius of convergence means you can move twice as far for the same loss of series precision.

~ Jay Daniel Fox