Math cleanup pass

- Use \mathit{...} to mark variables
- Use \operatorname{...} to mark operators/functions

Plus additional small fixes.

Resolves #839

Author: hollasch

books/RayTracingInOneWeekend.html

@@ -618,7 +618,7 @@
 This is commonly referred to as a _lerp_ between two values.
 A lerp is always of the form
 
-  $$ \text{blendedValue} = (1-a)\cdot\text{startValue} + a\cdot\text{endValue}, $$
+  $$ \mathit{blendedValue} = (1-a)\cdot\mathit{startValue} + a\cdot\mathit{endValue}, $$
 
 with $a$ going from zero to one.
 
@@ -784,13 +784,16 @@
 
 On the earth, this implies that the vector from the earth’s center to you points straight up. Let’s
 throw that into the code now, and shade it. We don’t have any lights or anything yet, so let’s just
-visualize the normals with a color map. A common trick used for visualizing normals (because it’s
-easy and somewhat intuitive to assume $\mathbf{n}$ is a unit length vector -- so each component is
-between -1 and 1) is to map each component to the interval from 0 to 1, and then map $(x, y, z)$ to
-$(\text{red}, \text{green}, \text{blue})$. For the normal, we need the hit point, not just whether
-we hit or not. We only have one sphere in the scene, and it's directly in front of the camera, so we
-won't worry about negative values of $t$ yet. We'll just assume the closest hit point
-(smallest $t$). These changes in the code let us compute and visualize $\mathbf{n}$:
+visualize the normals with a color map.
+A common trick used for visualizing normals (because it’s easy and somewhat intuitive to assume
+  $\mathbf{n}$ is a unit length vector -- so each component is between -1 and 1) is to map each
+  component to the interval from 0 to 1, and then map $(x, y, z)$ to $(\mathit{red}, \mathit{green},
+  \mathit{blue})$.
+For the normal, we need the hit point, not just whether we hit or not.
+We only have one sphere in the scene, and it's directly in front of the camera, so we won't worry
+  about negative values of $t$ yet.
+We'll just assume the closest hit point (smallest $t$).
+These changes in the code let us compute and visualize $\mathbf{n}$:
 
     ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ C++ highlight
     double hit_sphere(const point3& center, double radius, const ray& r) {

books/RayTracingTheNextWeek.html

@@ -1372,22 +1372,22 @@
 function `atan2()`, which takes any pair of numbers proportional to sine and cosine and returns the
 angle, we can pass in $x$ and $z$ (the $\sin(\theta)$ cancel) to solve for $\phi$:
 
-  $$ \phi = \text{atan2}(z, -x) $$
+  $$ \phi = \operatorname{atan2}(z, -x) $$
 
 `atan2()` returns values in the range $-\pi$ to $\pi$, but they go from 0 to $\pi$, then flip to
 $-\pi$ and proceed back to zero. While this is mathematically correct, we want $u$ to range from $0$
 to $1$, not from $0$ to $1/2$ and then from $-1/2$ to $0$. Fortunately,
 
-  $$ \text{atan2}(a,b) = \text{atan2}(-a,-b) + \pi, $$
+  $$ \operatorname{atan2}(a,b) = \operatorname{atan2}(-a,-b) + \pi, $$
 
 and the second formulation yields values from $0$ continuously to $2\pi$. Thus, we can compute
 $\phi$ as
 
-  $$ \phi = \text{atan2}(-z, x) + \pi $$
+  $$ \phi = \operatorname{atan2}(-z, x) + \pi $$
 
 The derivation for $\theta$ is more straightforward:
 
-  $$ \theta = \text{acos}(-y) $$
+  $$ \theta = \arccos(-y) $$
 
 <div class='together'>
 So for a sphere, the $(u,v)$ coord computation is accomplished by a utility function that takes
@@ -2491,7 +2491,7 @@
 represents the normal vector. To get this, we just use the cross product of the two side vectors
 $\mathbf{u}$ and $\mathbf{v}$:
 
-  $$ \mathbf{n} = \text{unit_vector}(\mathbf{u} \times \mathbf{v}) $$
+  $$ \mathbf{n} = \operatorname{unit\_vector}(\mathbf{u} \times \mathbf{v}) $$
 
 The plane is defined as all points $(x,y,z)$ that satisfy the equation $Ax + By + Cz = D$. Well, we
 know that $\mathbf{Q}$ lies on the plane, so that's enough to solve for $D$:
@@ -3658,7 +3658,7 @@
 As the ray passes through the volume, it may scatter at any point. The denser the volume, the more
 likely that is. The probability that the ray scatters in any small distance $\Delta L$ is:
 
-  $$ \text{probability} = C \cdot \Delta L $$
+  $$ \mathit{probability} = C \cdot \Delta L $$
 
 where $C$ is proportional to the optical density of the volume. If you go through all the
 differential equations, for a random number you get a distance where the scattering occurs. If that