11/17/2010, 02:24 PM
(This post was last modified: 11/17/2010, 03:55 PM by sheldonison.)
Here is a contour plot for the \( z+\theta(z) \) contour for \( \text{usexp}_{\sqrt 2}(z) \), where usexp(0)~=5.767 (not normalized). The contour is \( \theta(z)+z \) at the real axis, where \( \text{sexp}_{\sqrt 2}(z)=\text{usexp}_{\sqrt 2}(z+\theta(z)) \).
The first red contour represents the contour where \( \Im(\text{usexp}(z))=\pi/\log(\sqrt 2) \).
This becomes sexp(z) between z=-3 and z=-2.
The next green contour represents \( \text{usexp}(z)=-\infty \to 0 \), This becomes sexp(z) between z=-2 and z=-1.
The second red contour represents \( \text{usexp}(z)=0 \to 1 \)
The second green contour represents \( \text{usexp}(z)=1 \to \sqrt 2 \)
- Sheldon
\( \theta(z)+z \) has a similar contour at \( \Im(z)=17.143 \), exactly in between, at 8.571i, theta(z) is real valued, with a very small amplitude (4E-25). The "newsexp" z+theta(z) maps the same contour but without this second contour, and instead decays to 0 as imag(z) goes to infinity.
The first red contour represents the contour where \( \Im(\text{usexp}(z))=\pi/\log(\sqrt 2) \).
This becomes sexp(z) between z=-3 and z=-2.
The next green contour represents \( \text{usexp}(z)=-\infty \to 0 \), This becomes sexp(z) between z=-2 and z=-1.
The second red contour represents \( \text{usexp}(z)=0 \to 1 \)
The second green contour represents \( \text{usexp}(z)=1 \to \sqrt 2 \)
- Sheldon
\( \theta(z)+z \) has a similar contour at \( \Im(z)=17.143 \), exactly in between, at 8.571i, theta(z) is real valued, with a very small amplitude (4E-25). The "newsexp" z+theta(z) maps the same contour but without this second contour, and instead decays to 0 as imag(z) goes to infinity.
