make pdf file generation work

quarto/ODEs/differential_equations.qmd

@@ -72,13 +72,11 @@ $$
 The author's apply this model to  flu statistics from Hong Kong where:
 
 
-$$
 \begin{align*}
 S(0) &= 7,900,000\\
 I(0) &= 10\\
 R(0) &= 0\\
 \end{align*}
-$$
 
 In `Julia` we define these, `N` to model the total population, and `u0` to be the proportions.
 
@@ -133,13 +131,11 @@ The plot shows steady decay, as there is no mixing of infected with others.
 Adding in the interaction requires a bit more work. We now have what is known as a *system* of equations:
 
 
-$$
 \begin{align*}
 \frac{ds}{dt} &= -b   \cdot s(t)  \cdot  i(t)\\
 \frac{di}{dt} &= b  \cdot  s(t)  \cdot  i(t) - k  \cdot  i(t)\\
 \frac{dr}{dt} &= k  \cdot  i(t)\\
 \end{align*}
-$$
 
 Systems of equations can be solved in a similar manner as a single ordinary differential equation, though adjustments are made to accommodate the multiple functions.
 
@@ -282,12 +278,10 @@ We now solve numerically the problem of a trajectory with a drag force from air
 The general model is:
 
 
-$$
 \begin{align*}
 x''(t) &= - W(t,x(t), x'(t), y(t), y'(t)) \cdot x'(t)\\
 y''(t) &= -g - W(t,x(t), x'(t), y(t), y'(t)) \cdot y'(t)\\
 \end{align*}
-$$
 
 with initial conditions: $x(0) = y(0) = 0$ and $x'(0) = v_0 \cos(\theta), y'(0) = v_0 \sin(\theta)$.
 

quarto/ODEs/euler.qmd

@@ -70,7 +70,6 @@ That is, if we stitched together pieces of the slope field, would we get a curve
 ```{julia}
 #| hold: true
 #| echo: false
-#| cache: true
 ## {{{euler_graph}}}
 function make_euler_graph(n)
     x, y = symbols("x, y")
@@ -241,17 +240,7 @@ It is more work for the computer, but not for us, and clearly a much better appr
 
 ## The Euler method
 
-
-```{julia}
-#| hold: true
-#| echo: false
-imgfile ="figures/euler.png"
-caption = """
-Figure from first publication of Euler's method. From [Gander and Wanner](http://www.unige.ch/~gander/Preprints/Ritz.pdf).
-"""
-
-ImageFile(:ODEs, imgfile, caption)
-```
+![Figure from first publication of Euler's method. From [Gander and Wanner](http://www.unige.ch/~gander/Preprints/Ritz.pdf).](./figures/euler.png)
 
 The name of our function reflects the [mathematician](https://en.wikipedia.org/wiki/Leonhard_Euler) associated with the iteration:
 
@@ -361,9 +350,13 @@ caption = """
 A child's bead game. What shape wire will produce the shortest time for a bed to slide from a top to the bottom?
 
 """
-ImageFile(:ODEs, imgfile, caption)
+#ImageFile(:ODEs, imgfile, caption)
+nothing
 ```
 
+![A child's bead game. What shape wire will produce the shortest time for a bed to slide from a top to the bottom?](./figures/bead-game.jpg)
+
+
 Restrict our attention to the $x$-$y$ plane, and consider a path, between the point $(0,A)$ and $(B,0)$. Let $y(x)$ be the distance from $A$, so $y(0)=0$ and at the end $y$ will be $A$.
 
 
@@ -378,16 +371,22 @@ caption = """
 As early as 1638, Galileo showed that an object falling along `AC` and then `CB` will fall faster than one traveling along `AB`, where `C` is on the arc of a circle.
 From the [History of Math Archive](http://www-history.mcs.st-and.ac.uk/HistTopics/Brachistochrone.html).
 """
-ImageFile(:ODEs, imgfile, caption)
+#ImageFile(:ODEs, imgfile, caption)
+nothing
 ```
 
+![As early as 1638, Galileo showed that an object falling along `AC`
+and then `CB` will fall faster than one traveling along `AB`, where
+`C` is on the arc of a circle.  From the [History of Math
+Archive](http://www-history.mcs.st-and.ac.uk/HistTopics/Brachistochrone.html).
+](./figures/galileo.png)
+
 This simulation also suggests that a curved path is better than the shorter straight one:
 
 
 ```{julia}
 #| hold: true
 #| echo: false
-#| cache: true
 ##{{{brach_graph}}}
 
 function brach(f, x0, vx0, y0, vy0, dt, n)
@@ -603,13 +602,11 @@ $$
 We can try the Euler method here. A simple approach might be this iteration scheme:
 
 
-$$
 \begin{align*}
 x_{n+1} &= x_n + h,\\
 u_{n+1} &= u_n + h v_n,\\
 v_{n+1} &= v_n - h \cdot g/l \cdot \sin(u_n).
 \end{align*}
-$$
 
 Here we need *two* initial conditions: one for the initial value $u(t_0)$ and the initial value of $u'(t_0)$. We have seen if we start at an angle $a$ and release the bob from rest, so $u'(0)=0$ we get a sinusoidal answer to the linearized model. What happens here? We let $a=1$, $L=5$ and $g=9.8$:
 

quarto/ODEs/odes.qmd

@@ -87,13 +87,11 @@ $$
 Again, we can integrate to get an answer for any value $t$:
 
 
-$$
 \begin{align*}
 x(t) - x(t_0) &= \int_{t_0}^t \frac{dv}{dt} dt     \\
 &= (v_0t + \frac{1}{2}a t^2 - at_0 t) |_{t_0}^t    \\
 &= (v_0 - at_0)(t - t_0) + \frac{1}{2} a (t^2 - t_0^2).
 \end{align*}
-$$
 
 There are three constants: the initial value for the independent variable, $t_0$, and the two initial values for the velocity and position, $v_0, x_0$.  Assuming $t_0 = 0$, we can simplify the above to get a formula familiar from introductory physics:
 
@@ -339,12 +337,10 @@ Differential equations are classified according to their type. Different types h
 The first-order initial value equations we have seen can be described generally by
 
 
-$$
 \begin{align*}
 y'(x) &= F(y,x),\\
 y(x_0) &= x_0.
 \end{align*}
-$$
 
 Special cases include:
 
@@ -615,9 +611,11 @@ imgfile = "figures/verrazano-narrows-bridge-anniversary-historic-photos-2.jpeg"
 caption = """
 The cables of an unloaded suspension bridge have a different shape than a loaded suspension bridge. As seen, the cables in this [figure](https://www.brownstoner.com/brooklyn-life/verrazano-narrows-bridge-anniversary-historic-photos/) would be modeled by a catenary.
 """
-ImageFile(:ODEs, imgfile, caption)
+# ImageFile(:ODEs, imgfile, caption)
+nothing
 ```
 
+![The cables of an unloaded suspension bridge have a different shape than a loaded suspension bridge. As seen, the cables in this [figure](https://www.brownstoner.com/brooklyn-life/verrazano-narrows-bridge-anniversary-historic-photos/) would be modeled by a catenary.](./figures/verrazano-narrows-bridge-anniversary-historic-photos-2.jpeg)
 ---
 
 
@@ -668,13 +666,11 @@ Though `y` is messy, it can be seen that the answer is a quadratic polynomial in
 In a resistive medium, there are drag forces at play. If this force is proportional to the velocity, say, with proportion $\gamma$, then the equations become:
 
 
-$$
 \begin{align*}
 x''(t) &= -\gamma x'(t), & \quad y''(t) &= -\gamma y'(t) -g, \\
 x(0) &= x_0, &\quad y(0) &= y_0,\\
 x'(0) &= v_0\cos(\alpha),&\quad y'(0) &= v_0 \sin(\alpha).
 \end{align*}
-$$
 
 We now attempt to solve these.