finally got docs compiling

brunobeltran · brunobeltran · commit f1a1d603a12f · 2020-03-20T17:34:14.000-07:00
diff --git a/lib/matplotlib/bezier.py b/lib/matplotlib/bezier.py
@@ -290,37 +290,43 @@ def arc_area(self):
                       &\hspace{1em}- \left( \sum_{j=0}^n P_j^{(1)} b_{j,n} \right)
                         \left( n \sum_{k=0}^{n-1} (P_{k+1}^{(0)} -
                                 P_{k}^{(0)}) b_{j,n} \right)
-                      dt
+                      dt,
 
-        Where :math:`b_{\nu, n}(t) = {n \choose \nu} t^\nu {(1 - t)}^{n-\nu}`
+        where :math:`b_{\nu, n}(t) = {n \choose \nu} t^\nu {(1 - t)}^{n-\nu}`
         is the :math:`\nu`'th Bernstein polynomial of degree :math:`n`.
 
         Grouping :math:`t^l(1-t)^m` terms together for each :math:`l`,
         :math:`m`, we get that the integrand becomes
 
         .. math::
 
-            &\sum_{j=0}^n \sum_{k=0}^{n-1}
+            \sum_{j=0}^n \sum_{k=0}^{n-1}
                 {n \choose j} {{n - 1} \choose k}
-                [P_j^{(0)} (P_{k+1}^{(1)} - P_{k}^{(1)})
-                    - P_j^{(1)} (P_{k+1}^{(0)} - P_{k}^{(0)})]
-                t^{j + k} {(1 - t)}^{2n - 1 - j - k}
-                \\
-            &= \sum_{j=0}^n \sum_{k=0}^{n-1}
+                &\left[P_j^{(0)} (P_{k+1}^{(1)} - P_{k}^{(1)})
+                    - P_j^{(1)} (P_{k+1}^{(0)} - P_{k}^{(0)})\right] \\
+                &\hspace{1em}\times{}t^{j + k} {(1 - t)}^{2n - 1 - j - k}
+
+        or just
+
+        .. math::
+
+            \sum_{j=0}^n \sum_{k=0}^{n-1}
                 \frac{{n \choose j} {{n - 1} \choose k}}
                         {{{2n - 1} \choose {j+k}}}
                 [P_j^{(0)} (P_{k+1}^{(1)} - P_{k}^{(1)})
                     - P_j^{(1)} (P_{k+1}^{(0)} - P_{k}^{(0)})]
-                b_{j+k,2n-1}(t)
+                b_{j+k,2n-1}(t).
 
         Interchanging sum and integral, and using the fact that :math:`\int_0^1
         b_{\nu, n}(t) dt = \frac{1}{n + 1}`, we conclude that the
-        original integral (:math:`\frac{1}{2}\int_0^1 B(t) \cdot n(t) dt`) can
+        original integral  can
         simply be written as
 
         .. math::
 
-            \frac{1}{4}\sum_{j=0}^n \sum_{k=0}^{n-1}
+            \frac{1}{2}&\int_0^1 B(t) \cdot n(t) dt
+            \\
+            &= \frac{1}{4}\sum_{j=0}^n \sum_{k=0}^{n-1}
               \frac{{n \choose j} {{n - 1} \choose k}}
                     {{{2n - 1} \choose {j+k}}}
               [P_j^{(0)} (P_{k+1}^{(1)} - P_{k}^{(1)})
