Use :math: in a few docstrings in the cones module.

diff --git a/dunshire/cones.py b/dunshire/cones.py
index bf873dab1c09021873a27b8cfd9d82b3523e9e1a..50c57e747952cd207005d82ef25b6817e29bef61 100644 (file)
--- a/dunshire/cones.py
+++ b/dunshire/cones.py
@@ -402,7 +402,7 @@ class IceCream(SymmetricCone):
 
 
     def ball_radius(self, point):
-        """
+        r"""
         Return the radius of a ball around ``point`` in this cone.
 
         Since a radius cannot be negative, the ``point`` must be
@@ -411,12 +411,12 @@ class IceCream(SymmetricCone):
         The minimum distance from ``point`` to the complement of this
         cone will always occur at its projection onto that set. It is
         not hard to see that the projection is at a "down and out" angle
-        of ``pi/4`` towards the outside of the cone. If one draws a
+        of :math:`\pi/4` towards the outside of the cone. If one draws a
         right triangle involving the ``point`` and that projection, it
         becomes clear that the distance between ``point`` and its
-        projection is a factor of ``1/sqrt(2)`` times the "horizontal"
+        projection is a factor of :math:`1/\sqrt(2)` times the "horizontal"
         distance from ``point`` to boundary of this cone. For simplicity
-        we take ``1/2`` instead.
+        we take :math:`1/2` instead.
 
         Parameters
         ----------
@@ -446,16 +446,17 @@ class IceCream(SymmetricCone):
         Examples
         --------
 
-        The height of ``x`` is one (its first coordinate), and so the
-        radius of the circle obtained from a height-one cross section is
-        also one. Note that the last two coordinates of ``x`` are half
-        of the way to the boundary of the cone, and in the direction of
-        a 30-60-90 triangle. If one follows those coordinates, they hit
-        at ``(1, sqrt(3)/2, 1/2)`` having unit norm. Thus the
-        "horizontal" distance to the boundary of the cone is ``1 -
-        norm(x)``, which simplifies to ``1/2``. And rather than involve
-        a square root, we divide by two for a final safe radius of
-        ``1/4``.
+        The height of ``x`` below is one (its first coordinate), and so
+        the radius of the circle obtained from a height-one cross
+        section is also one. Note that the last two coordinates of ``x``
+        are half of the way to the boundary of the cone, and in the
+        direction of a 30-60-90 triangle. If one follows those
+        coordinates, they hit at :math:`\left(1, \frac{\sqrt(3)}{2},
+        \frac{1}{2}\right)` having unit norm. Thus the "horizontal"
+        distance to the boundary of the cone is :math:`1 - \left\lVert x
+        \right\rVert`, which simplifies to :math:`1/2`. And rather than
+        involve a square root, we divide by two for a final safe radius
+        of :math:`1/4`.
 
             >>> from math import sqrt
             >>> K = IceCream(3)
@@ -749,11 +750,10 @@ class CartesianProduct(SymmetricCone):
 
     def cvxopt_dims(self):
         """
-        Return a dictionary of dimensions corresponding to the factors
-        of this cartesian product. The format of this dictionary is
-        described in the CVXOPT user's guide:
-
-          http://cvxopt.org/userguide/coneprog.html#linear-cone-programs
+        Return a dictionary of dimensions corresponding to the
+        factors of this cartesian product. The format of this dictionary
+        is described in the `CVXOPT user's guide
+        <http://cvxopt.org/userguide/coneprog.html#linear-cone-programs>`_.
 
         Returns
         -------