Change \textit to \mathit

Author: ymherklotz

books/RayTracingTheRestOfYourLife.html

@@ -388,7 +388,7 @@
 Our Buffon Needle example is a way of calculating $\pi$ by solving for the ratio of the area of the
 circle and the area of the inscribing square:
 
-    $$ \frac{\text{area}(\textit{circle})}{\text{area}(\textit{square})} = \frac{\pi}{4} $$
+    $$ \frac{\text{area}(\mathit{circle})}{\text{area}(\mathit{square})} = \frac{\pi}{4} $$
 
 We picked a bunch of random points in the inscribing square and counted the fraction of them that
 were also in the unit circle. This fraction was an estimate that tended toward $\frac{\pi}{4}$ as
@@ -397,14 +397,14 @@
 $\frac{\pi}{4}$, and we know that the area of a inscribing square is $4r^2$, so we could then use
 those two quanties to get the area of a circle:
 
-    $$ \frac{\text{area}(\textit{circle})}{\text{area}(\textit{square})} = \frac{\pi}{4} $$
-    $$ \frac{\text{area}(\textit{circle})}{(2r)^2} = \frac{\pi}{4} $$
-    $$ \text{area}(\textit{circle}) = \frac{\pi}{4} 4r^2 $$
-    $$ \text{area}(\textit{circle}) = \pi r^2 $$
+    $$ \frac{\text{area}(\mathit{circle})}{\text{area}(\mathit{square})} = \frac{\pi}{4} $$
+    $$ \frac{\text{area}(\mathit{circle})}{(2r)^2} = \frac{\pi}{4} $$
+    $$ \text{area}(\mathit{circle}) = \frac{\pi}{4} 4r^2 $$
+    $$ \text{area}(\mathit{circle}) = \pi r^2 $$
 
 We choose a circle with radius $r = 1$ and get:
 
-    $$ \text{area}(\textit{circle}) = \pi $$
+    $$ \text{area}(\mathit{circle}) = \pi $$
 
 Our work above is equally valid as a means to solve for $pi$ as it is a means to solve for the area
 of a circle:
@@ -1340,12 +1340,12 @@
 in the _LMS color space_ (long, medium, short).
 
 If the light does scatter, it will have a directional distribution that we can describe as a PDF
-over solid angle. I will refer to this as its _scattering PDF_: $\textit{pScatter}()$. The scattering PDF
-will vary with the outgoing direction: $\textit{pScatter}(\omega_o)$. The scattering PDF can also vary with
-_incident direction_: $\textit{pScatter}(\omega_i, \omega_o)$. You can see this varying with incident
+over solid angle. I will refer to this as its _scattering PDF_: $\mathit{pScatter}()$. The scattering PDF
+will vary with the outgoing direction: $\mathit{pScatter}(\omega_o)$. The scattering PDF can also vary with
+_incident direction_: $\mathit{pScatter}(\omega_i, \omega_o)$. You can see this varying with incident
 direction when you look at reflections off a road -- they become mirror-like as your viewing angle
 (incident angle) approaches grazing. Lastly, the scattering PDF can also depend on the scattering
-position: $\textit{pScatter}(\textbf{x}, \omega_i, \omega_o)$. The $\textbf{x}$ is just a vector representing
+position: $\mathit{pScatter}(\textbf{x}, \omega_i, \omega_o)$. The $\textbf{x}$ is just a vector representing
 the scattering position: $\textbf{x} = (x, y, z)$. The Albedo of an object can also depend on these
 quantities: $A(\textbf{x}, \omega_i, \omega_o)$.
 
@@ -1355,15 +1355,15 @@
 
     $$ \text{Color}_o(\textbf{x}, \omega_o) = \int_{\omega_i}
         A(\textbf{x}, \omega_i, \omega_o) \cdot
-        \textit{pScatter}(\textbf{x}, \omega_i, \omega_o) \cdot
+        \mathit{pScatter}(\textbf{x}, \omega_i, \omega_o) \cdot
         \text{Color}_i(\textbf{x}, \omega_i) $$
 
 We've added a $\text{Color}_i$ term. The scattering PDF and the albedo at the surface of an object
 are acting as filters to the light that is shining on that point. So we need to solve for the light
 that is shining on that point. This is a recursive algorithm, and is the reason our `ray_color`
 function returns the color of the current object multiplied by the color of the next ray.
 
-Note that $A()$, $\textit{pScatter}()$, and $\text{Color}_i()$ may all depend on the wavelength of the light,
+Note that $A()$, $\mathit{pScatter}()$, and $\text{Color}_i()$ may all depend on the wavelength of the light,
 $\lambda$, but I've left it out of the equation because it's complicated enough as it is.
 </div>
 
@@ -1374,7 +1374,7 @@
 If we apply the Monte Carlo basic formula we get the following statistical estimate:
 
     $$ \text{Color}_o = \frac{A(\textbf{x}, \omega_i, \omega_o) \cdot
-        \textit{pScatter}(\textbf{x}, \omega_i, \omega_o) \cdot
+        \mathit{pScatter}(\textbf{x}, \omega_i, \omega_o) \cdot
         \text{Color}_i(\textbf{x}, \omega_i)}
         {p(\textbf{x}, \omega_o)} $$
 
@@ -1383,10 +1383,10 @@
 </div>
 
 For a Lambertian surface we already implicitly implemented this formula for the special case where
-$p()$ is a cosine density. The $\textit{pScatter}()$ of a Lambertian surface is proportional to
+$p()$ is a cosine density. The $\mathit{pScatter}()$ of a Lambertian surface is proportional to
 $\cos(\theta_o)$, where $\theta_o$ is the angle relative to the surface normal. All two dimensional
-PDFs need to integrate to one over the whole surface (remember that $\textit{pScatter}()$ is a PDF). We set
-$\textit{pScatter}(\theta_o < 0) = 0$ so that we don't scatter below the horizon. The integral of
+PDFs need to integrate to one over the whole surface (remember that $\mathit{pScatter}()$ is a PDF). We set
+$\mathit{pScatter}(\theta_o < 0) = 0$ so that we don't scatter below the horizon. The integral of
 $\cos(\theta_o)$ over the hemisphere is $\pi$.
 
 <div class='together'>
@@ -1404,13 +1404,13 @@
 If we have a scattering PDF that is $\cos(\theta_o)$ over the hemisphere, then we need to normalize
 by $\pi$. So for a Lambertian surface the scattering PDF is:
 
-    $$ \textit{pScatter}(\omega_o) = \frac{\cos(\theta_o)}{\pi} $$
+    $$ \mathit{pScatter}(\omega_o) = \frac{\cos(\theta_o)}{\pi} $$
 </div>
 
 <div class='together'>
 We'll assume that the PDF is equal to the scattering PDF:
 
-    $$ p(\omega_o) = \textit{pScatter}() = \frac{\cos(\theta_o)}{\pi} $$
+    $$ p(\omega_o) = \mathit{pScatter}() = \frac{\cos(\theta_o)}{\pi} $$
 
 The numerator and denominator cancel out, and we get:
 
@@ -1428,7 +1428,7 @@
 _Bidirectional Reflectance Distribution Function_ (BRDF). It relates pretty simply to our terms:
 
     $$ BRDF(\omega_i, \omega_o) = \frac{A(\textbf{x}, \omega_i, \omega_o) \cdot
-        \textit{pScatter}(\textbf{x}, \omega_i, \omega_o)}{\cos(\theta_o)} $$
+        \mathit{pScatter}(\textbf{x}, \omega_i, \omega_o)}{\cos(\theta_o)} $$
 
 So for a Lambertian surface for example, $BRDF = A / \pi$. Translation between our terms and BRDF is
 easy. For participating media (volumes), our albedo is usually called the _scattering albedo_, and
@@ -1446,26 +1446,26 @@
 <div class='together'>
 Our goal over the next two chapters is to instrument our program to send a bunch of extra rays
 toward light sources so that our picture is less noisy. Let’s assume we can send a bunch of rays
-toward the light source using a PDF $\textit{pLight}(\omega_o)$. Let’s also assume we have a PDF related to
-$\textit{pScatter}$, and let’s call that $\textit{pSurface}(\omega_o)$. A great thing about PDFs is that you can just
+toward the light source using a PDF $\mathit{pLight}(\omega_o)$. Let’s also assume we have a PDF related to
+$\mathit{pScatter}$, and let’s call that $\mathit{pSurface}(\omega_o)$. A great thing about PDFs is that you can just
 use linear mixtures of them to form mixture densities that are also PDFs. For example, the simplest
 would be:
 
-  $$ p(\omega_o) = \frac{1}{2} \textit{pSurface}(\omega_o) +  \frac{1}{2} \textit{pLight}(\omega_o)$$
+  $$ p(\omega_o) = \frac{1}{2} \mathit{pSurface}(\omega_o) +  \frac{1}{2} \mathit{pLight}(\omega_o)$$
 </div>
 
 As long as the weights are positive and add up to one, any such mixture of PDFs is a PDF. Remember,
 we can use any PDF: _all PDFs eventually converge to the correct answer_. So, the game is to figure
 out how to make the PDF larger where the product
 
-    $$ \textit{pScatter}(\textbf{x}, \omega_i, \omega_o) \cdot \text{Color}_i(\textbf{x}, \omega_i) $$
+    $$ \mathit{pScatter}(\textbf{x}, \omega_i, \omega_o) \cdot \text{Color}_i(\textbf{x}, \omega_i) $$
 
 is largest. For diffuse surfaces, this is mainly a matter of guessing where
 $\text{Color}_i(\textbf{x}, \omega_i)$ is largest. Which is equivalent to guessing where the most
 light is coming from.
 
-For a mirror, $\textit{pScatter}()$ is huge only near one direction, so $\textit{pScatter}()$ matters a lot more. In
-fact, most renderers just make mirrors a special case, and make the $\textit{pScatter}()/p()$ implicit -- our
+For a mirror, $\mathit{pScatter}()$ is huge only near one direction, so $\mathit{pScatter}()$ matters a lot more. In
+fact, most renderers just make mirrors a special case, and make the $\mathit{pScatter}()/p()$ implicit -- our
 code currently does that.
 
 
@@ -1674,7 +1674,7 @@
 material, but Lambertian requires a $\cos(\theta_o)$ sampling pattern. So any differences that we
 see is because we are using a slightly different material. We're going to uniformly scatter random
 directions from the hemisphere and perfectly match the PDF, given by
-$\textit{pScatter}(\omega_o) = p(\omega_o) = \frac{1}{2\pi}$.
+$\mathit{pScatter}(\omega_o) = p(\omega_o) = \frac{1}{2\pi}$.
 
     ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ C++
     class lambertian : public material {
@@ -2416,7 +2416,7 @@
 </div>
 
 <div class='together'>
-The details of how this is done under the hood varies for $\textit{pSurface}$ and $\textit{pLight}$, but that is
+The details of how this is done under the hood varies for $\mathit{pSurface}$ and $\mathit{pLight}$, but that is
 exactly what class hierarchies were invented for! It’s never obvious what goes in an abstract class,
 so my approach is to be greedy and hope a minimal interface works, and for `pdf` this implies:
 
@@ -2735,13 +2735,13 @@
 mixtures of any PDFs to form mixture densities that are also PDFs. Any weighted average of PDFs is
 also a PDF. As long as the weights are positive and add up to any one, we have a new PDF.
 
-  $$ \textit{pMixture}() = w_0 p_0() + w_1 p_1() + w_2 p_2() + \ldots + w_{n-1} p_{n-1}() $$
+  $$ \mathit{pMixture}() = w_0 p_0() + w_1 p_1() + w_2 p_2() + \ldots + w_{n-1} p_{n-1}() $$
 
   $$ 1 = w_0 + w_1 + w_2 + \ldots + w_{n-1} $$
 
 For example, we could just average the two densities:
 
-  $$ \textit{pMixture}(\omega_o) = \frac{1}{2} \textit{pSurface}(\omega_o) +  \frac{1}{2} \textit{pLight}(\omega_o)$$
+  $$ \mathit{pMixture}(\omega_o) = \frac{1}{2} \mathit{pSurface}(\omega_o) +  \frac{1}{2} \mathit{pLight}(\omega_o)$$
 
 <div class='together'>
 How would we instrument our code to do that? There is a very important detail that makes this not
@@ -2754,7 +2754,7 @@
         pick direction according to pLight
     ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
 
-But solving for the PDF value of $\textit{pMixture}$ is slightly more subtle. We can't just
+But solving for the PDF value of $\mathit{pMixture}$ is slightly more subtle. We can't just
 
     ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
     if (direction is from pSurface)
@@ -2769,8 +2769,8 @@
 correct, we would have to solve backwards to figure which PDF the direction could come from. Which
 honestly sounds like a nightmare, but fortunately we don't need to do that. There are some
 directions that both PDFs could have generated. For example, a direction toward the light could have
-been generated by either $\textit{pLight}$ _or_ $\textit{pSurface}$. It is sufficient for us to solve for the pdf
-value of $\textit{pSurface}$ and of $\textit{pLight}$ for a random direction and then take the PDF mixture weights to
+been generated by either $\mathit{pLight}$ _or_ $\mathit{pSurface}$. It is sufficient for us to solve for the pdf
+value of $\mathit{pSurface}$ and of $\mathit{pLight}$ for a random direction and then take the PDF mixture weights to
 solve for the total PDF value for that direction. The mixture density class is actually pretty
 straightforward: