em dash; sentence case

quarto/differentiable_vector_calculus/scalar_functions_applications.qmd

@@ -918,7 +918,7 @@ zs = fₗ.(xs, ys)
 scatter3d!(xs, ys, zs)
 ```
 
-A contour plot also shows that some - and only one - extrema happens on the interior:
+A contour plot also shows that some---and only one---extrema happens on the interior:
 
 
 ```{julia}
@@ -967,10 +967,10 @@ We confirm this by looking at the Hessian and noting $H_{11} > 0$:
 Hₛ = subs.(hessian(exₛ, [x,y]), x=>xstarₛ[x], y=>xstarₛ[y])
 ```
 
-As it occurs at $(\bar{x}, \bar{y})$ where $\bar{x} = (x_1 + x_2 + x_3)/3$ and $\bar{y} = (y_1+y_2+y_3)/3$ - the averages of the three values - the critical point is an interior point of the triangle.
+As it occurs at $(\bar{x}, \bar{y})$ where $\bar{x} = (x_1 + x_2 + x_3)/3$ and $\bar{y} = (y_1+y_2+y_3)/3$---the averages of the three values---the critical point is an interior point of the triangle.
 
 
-As mentioned by Strang, the real problem is to minimize $d_1 + d_2 + d_3$. A direct approach with `SymPy` - just replacing `d2` above with the square root fails. Consider instead the gradient of $d_1$, say. To avoid square roots, this is taken implicitly from $d_1^2$:
+As mentioned by Strang, the real problem is to minimize $d_1 + d_2 + d_3$. A direct approach with `SymPy`---just replacing `d2` above with the square root fails. Consider instead the gradient of $d_1$, say. To avoid square roots, this is taken implicitly from $d_1^2$:
 
 
 $$
@@ -1016,7 +1016,7 @@ psₛₗ = [a*u for (a,u) in zip(asₛ₁, usₛ)]
 plot!(polygon(psₛₗ)...)
 ```
 
-Let's see where the minimum distance point is by constructing a plot. The minimum must be on the boundary, as the only point where the gradient vanishes is the origin, not in the triangle. The plot of the triangle has a contour plot of the distance function, so we see clearly that the minimum happens at the point `[0.5, -0.866025]`. On this plot, we drew the gradient at some points along the boundary. The gradient points in the direction of greatest increase - away from the minimum. That the gradient vectors have a non-zero projection onto the edges of the triangle in a direction pointing away from the point indicates that the function `d` would increase if moved along the boundary in that direction, as indeed it does.
+Let's see where the minimum distance point is by constructing a plot. The minimum must be on the boundary, as the only point where the gradient vanishes is the origin, not in the triangle. The plot of the triangle has a contour plot of the distance function, so we see clearly that the minimum happens at the point `[0.5, -0.866025]`. On this plot, we drew the gradient at some points along the boundary. The gradient points in the direction of greatest increase---away from the minimum. That the gradient vectors have a non-zero projection onto the edges of the triangle in a direction pointing away from the point indicates that the function `d` would increase if moved along the boundary in that direction, as indeed it does.
 
 
 ```{julia}
@@ -1064,7 +1064,7 @@ The smallest value is when $t=0$ or $t=1$, so at one of the points, as `li` is d
 ##### Example: least squares
 
 
-We know that two points determine a line. What happens when there are more than two points? This is common in statistics where a bivariate data set (pairs of points $(x,y)$) are summarized through a linear model $\mu_{y|x} = \alpha + \beta x$, That is the average value for $y$ given a particular $x$ value is given through the equation of a line. The data is used to identify what the slope and intercept are for this line. We consider a simple case - $3$ points. The case of $n \geq 3$ being similar.
+We know that two points determine a line. What happens when there are more than two points? This is common in statistics where a bivariate data set (pairs of points $(x,y)$) are summarized through a linear model $\mu_{y|x} = \alpha + \beta x$, That is the average value for $y$ given a particular $x$ value is given through the equation of a line. The data is used to identify what the slope and intercept are for this line. We consider a simple case---$3$ points. The case of $n \geq 3$ being similar.
 
 
 We have a line $l(x) = \alpha + \beta(x)$ and three points $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$. Unless these three points *happen* to be collinear, they can't possibly all lie on the same line. So to *approximate* a relationship by a line requires some inexactness. One measure of inexactness is the *vertical* distance to the line:
@@ -1118,7 +1118,7 @@ As found, the formulas aren't pretty. If $x_1 + x_2 + x_3 = 0$ they simplify. Fo
 subs(outₗₛ[β], sum(xₗₛ) => 0)
 ```
 
-Let $\vec{x} = \langle x_1, x_2, x_3 \rangle$ and $\vec{y} = \langle y_1, y_2, y_3 \rangle$ this is simply $(\vec{x} \cdot \vec{y})/(\vec{x}\cdot \vec{x})$, a formula that will generalize to $n > 3$. The assumption is not a restriction - it comes about by subtracting the mean, $\bar{x} = (x_1 + x_2 + x_3)/3$, from each $x$ term (and similarly subtract $\bar{y}$ from each $y$ term). A process called "centering."
+Let $\vec{x} = \langle x_1, x_2, x_3 \rangle$ and $\vec{y} = \langle y_1, y_2, y_3 \rangle$ this is simply $(\vec{x} \cdot \vec{y})/(\vec{x}\cdot \vec{x})$, a formula that will generalize to $n > 3$. The assumption is not a restriction---it comes about by subtracting the mean, $\bar{x} = (x_1 + x_2 + x_3)/3$, from each $x$ term (and similarly subtract $\bar{y}$ from each $y$ term). A process called "centering."
 
 
 With this observation, the formulas can be re-expressed through:
@@ -1587,7 +1587,7 @@ $$
 G(\epsilon_1, \epsilon_2) = L.
 $$
 
-Now, Lagrange's method can be employed. This will be fruitful - even though we know the answer - it being $\epsilon_1 = \epsilon_2 = 0$!
+Now, Lagrange's method can be employed. This will be fruitful---even though we know the answer---it being $\epsilon_1 = \epsilon_2 = 0$!
 
 
 Forging ahead, we compute $\nabla{F}$ and $\lambda \nabla{G}$ and set $\epsilon_1 = \epsilon_2 = 0$ where the two  are equal. This will lead to a description of $y$ in terms of $y'$.