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better equation
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brsr committed Jan 7, 2019
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2 changes: 1 addition & 1 deletion docs/euclidian_spherical_maps.aux
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25 changes: 3 additions & 22 deletions docs/euclidian_spherical_maps.tex
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Expand Up @@ -517,32 +517,13 @@ \section{Maps between polygons and disks in the Euclidean plane}
\hat{\mathbf v} = [x, y, \sqrt{1-x^2-y^2}].
\end{equation}
\subsection{Conformal}
The Schwarz-Christoffel mapping is a conformal mapping from a polygon in the
plan to the complex upper half-plane. Let $\{ \zeta \in \mathbb{C}:
\operatorname{Im} \zeta > 0 \}$ be a point in the complex upper half-plane. Let
$\alpha,\beta,\gamma, \ldots$ be the interior angles of the polygon, and $a, b,
c, \ldots$ be real numbers along the boundary of the upper half-plane. Then the
map from the upper half-plane to the polygon is:
\begin{equation}
f(\zeta) = \int^\zeta \frac{K}{(w-a)^{1-(\alpha/\pi)} (w-b)^{1-(\beta/\pi)}
(w-c)^{1-(\gamma/\pi)} \cdots} \,\mathrm{d}w
\end{equation}
or
\begin{equation}
f(z) = \int_0^z \prod_{k=1}^n (\zeta - z_k)^{\alpha_k-1} d\zeta
\end{equation}
where $K$ is an arbitrary constant. Like the Schwarz triangle map above,
Like the Schwarz triangle map above,
computing this integral usually results in a hypergeometric function,
which is not particularly convenient. For squares it results in an elliptic function, which is slightly more convenient.\cite{fong16}

The upper half-plane can be mapped to a unit disk by various Möbius
transformations. A common choice is this pair of equations,
where $z$ is a point on the disk and $\zeta$ is a point on the upper half-plane:
\begin{equation}\begin{split}
z &=\frac{i\zeta+1}{\zeta+i} \\
\zeta &= \frac{1 - i z}{z - i}
\end{split}\end{equation}
This maps $\zeta = i$ to $z = 0$ and $\zeta = 0$ to $z = -i$.
which is not particularly convenient. For squares it results in an elliptic
function, which is slightly more convenient.\cite{fong16}

\subsection{Naive Slerp}

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