Change symbol from n_0 to n_{s_0}

docs/Proportional-Interval-Stamp/Proportional-Interval-Stamp.mdx

@@ -131,7 +131,7 @@ I'll soon explain it with precise mathematical formulas, so don't worry about it
 
 Proportional interval has another benefit: Strokes have more consistent appearance!
 Watch the start and end points of the fixed interval stroke.
-Because the stroke radius drops to zero at these points, these dots are too sparser to keep a continuous feeling.
+Because the stroke radius drops to zero at these points, these dots are too sparser to keep a continuous apperance.
 In contrast, the proportional interval is consistent all along the stroke.
 Actually, it's a [fractal](https://en.wikipedia.org/wiki/Fractal) when a radius comes to zero.
 
@@ -202,7 +202,7 @@ $$
 As $x = L$, we know the total stamp number on the edge, remind that $\cos\theta L = r_0-r_1$:
 
 $$
-\tag{3} n(L) = \frac{L}{\eta (r_0-r_1)}\ln \frac{r_0}{r_1}
+\tag{3} n_L = \frac{L}{\eta (r_0-r_1)}\ln \frac{r_0}{r_1}
 $$
 
 We will soon use the formula (1)(2)(3) in our code.
@@ -219,20 +219,21 @@ Label the two roots with $x_1$ and $x_2$.
 ![locate stamp](./locate-stamp.png)
 
 To calculate the nearest stamp point (the black dot in the figure) between $x_1$ and $x_2$,
-we need to compute the number of stamps from $x_1$ to the polyline's first vertex, which is also called stamp index at $x_1$.
+we need to compute the number of stamps from $x_1$ to the polyline's first vertex, which is also called stamp index (or stamp count) at $x_1$.
 According to formula (1), on the current edge, the number of stamps from vertex0 to $x_1$ is $n(x_1)$.
-Label the stamp index of vertex0 as $n_0$, and label vertex1's as $n_1$.
+Label the stamp index of vertex0 as $n_{s_0}$, and label vertex1's as $n_{s_1}$.
+(The symbol $s$ represents the arc from the first to current vertex).
 
 ![store](./store-n.png)
 
-Therefore, the stamp index of $x_1$ is $n_0 + n(x_1)$.
+Therefore, the stamp index of $x_1$ is $n_{s_0} + n(x_1)$.
 Because we place footprints at the points with integer stamp index,
-the ceiling of $n_0 + n(x_1)$ is the nearest stamp's index,
-denoted as $n_\mathrm{nearest} = \lceil n_0 + n(x_1) \rceil$.
-Replace it into the formula (2) to get its position $x_{\mathrm{nearest}} = x(\lceil n_0 + n(x_1) \rceil)$,
+the ceiling of $n_{s_0} + n(x_1)$ is the nearest stamp's index,
+denoted as $n_\mathrm{nearest} = \lceil n_{s_0} + n(x_1) \rceil$.
+Replace it into the formula (2) to get its position $x_{\mathrm{nearest}} = x(\lceil n_{s_0} + n(x_1) \rceil)$,
 and all other positions $x(n_\mathrm{nearest} + 1)$, $x(n_\mathrm{nearest} + 2) \dots$
 
-Additionally, to determine $n_0$ and $n_1$, we can compute the prefix sum of the $n(L)$ in formula (3) over all edges.
+Additionally, to determine $n_{s_0}$ and $n_{s_1}$, we can compute the prefix sum of the $n_L$ in formula (3) over all edges.
 Put them into vertex data and pass them into fragment shader, exactly same as the value `length` in the Stamp section.
 
 ## Implementation
@@ -249,8 +250,8 @@ This will bring numeric errors, which we surely want to avoid.
 A simple solution is to add a small number to the radii, as demoed in the code above.
 For a more rigorous approach, you can implement checks to prevent $r_0/r_1$ in the logarithm from becoming zero or excessively large.
 
-Also, it's worth noting that if calculating the prefix sum value of $n(L)$ with CPU,
-maybe you want to push $\eta$ times $n(L)$, $\eta n(L)$ instead of $n(L)$ itself into vertex buffer.
+Also, it's worth noting that if calculating the prefix sum value of $n_L$ with CPU,
+maybe you want to push $\eta$ times $n_L$, $\eta n_L$ instead of $n_L$ itself into vertex buffer.
 Because $\eta$ is a value changed more frequently, you won't want to recompute the prefix sum every time it changes.
 
 ## Proof of properties