Searching for an asymptotic to exp[0.5]

[sheldonison]

\( \text{dhalfi}(y,z)=\frac{d}{dz} (\exp^{0.5}(y+zi)+\exp^{0.5}(y-zi)) \;\;\; mi(y) = \text{dhalfi}^{-1}(y,z)=0 \)
....Each z=pi i corresponds to half way around the circle. For example, for y=2, we get mi(2)~=5.65pi, so we wrap the approximation around the circle 5.65 times, for a radius of exp(2). This limit converges rapidly as y increases.

\( a_n = \lim_{y \to \infty} \frac{1}{2\pi}
\int_{-mi(y)}^{+mi(y)} \exp(\exp^{0.5} (y+iz) - n(y+iz))\;dz \)

Here are a couple of pictures showing how the equations for a_n converge. The top picture shows the optimal number of Pi steps for each value of y; this would be exactly mi(y)/Pi. The black line in the top picture shows how many pi steps are required to match the 1/x laurent error term; this is useful to understand which of the error terms are dominant in the approximation (more later).

The bottom picture shows the exponential improvement in the accuracy of the approximations for a_n, via the black error term, at the same time as the function is rapidly growing (purple). This is the graph for the a0 coeffient precision required for the calculations (purple-black) to get the optimal precision possible, in black.