diff --git a/figures.ipynb b/figures.ipynb
index 2f8ff0d00f9e446305fcc07d3c884f40644caec8..a7b1c466b7fc1bb2f46783ec918cb1ae0944acf5 100644
--- a/figures.ipynb
+++ b/figures.ipynb
@@ -378,7 +378,7 @@
     "axs.text(\n",
     "    -0.1,\n",
     "    0.5,\n",
-    "    (r\"$\\mathrm{time}$\" + \"\\n\" + \"$\\longleftarrow$\"),\n",
+    "    (r\"$\\mathrm{iterations}$\" + \"\\n\" + \"$\\longleftarrow$\"),\n",
     "    transform=axs.transAxes,\n",
     "    horizontalalignment=\"center\",\n",
     "    verticalalignment=\"center\",\n",
diff --git a/figures/algo.pdf b/figures/algo.pdf
index 0f1821b3ffd3a2d7f1f1d9174cd21f3efb086ca3..675243d8dbfc3749101ac8481870e6a514b5a5a2 100644
Binary files a/figures/algo.pdf and b/figures/algo.pdf differ
diff --git a/paper.md b/paper.md
index 643db28bec58bcaf6ca833fb07fe5f42963f7854..67a292764d5ef7a5a8b0d3b5808f695f3c3e4603 100755
--- a/paper.md
+++ b/paper.md
@@ -40,7 +40,7 @@ One of the most significant complications here is to parallelize this algorithm,
 We start by calculating the two boundary points.
 Two consecutive existing data points (black) $\{x_i, y_i\}$ define an interval.
 Each interval has a loss associated with it that can be calculated from the points inside the interval $L_{i,i+1}(x_i, x_{i+1}, y_i, y_{i+1})$.
-At each time step the interval with the largest loss is indicated (red), with its corresponding candidate point (green) picked in the middle of the interval.
+At each iteration the interval with the largest loss is indicated (red), with its corresponding candidate point (green) picked in the middle of the interval.
 The loss function in this example is the curvature loss.
 ](figures/algo.pdf){#fig:algo}
 
@@ -201,12 +201,14 @@ By adding the two loss functions, we can combine the 3D area loss to exploit int
 Inspired by a method commonly employed in digital cartography for coastline simplification, Visvalingam's algorithm, we construct a loss function that does its reverse.[@visvalingam1990douglas]
 Here, at each point (ignoring the boundary points), we compute the effective area associated with its triangle, see Fig. @fig:line_loss(b).
 The loss then becomes the average area of two adjacent traingles.
+By Taylor expanding $f$ around $x$ it can be shown that the area of the triangles relates to the contributions of the second derivative.
+We can generalize this loss to $N$ dimensions
 
 ![Line loss visualization.
 We start with 6 points (a) on the function (grey).
 Ignoring the endpoints, the effective area of each point is determined by its associated triangle (b).
 The loss of each interval can be computed by taking the average area of the adjacent triangles.
-Subplots (c), (d), and (e) show the subsequent time steps following (b).](figures/line_loss.pdf){#fig:line_loss}
+Subplots (c), (d), and (e) show the subsequent interations following (b).](figures/line_loss.pdf){#fig:line_loss}
 
 <!-- https://bost.ocks.org/mike/simplify/ -->