make pdf file generation work
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@@ -70,7 +70,6 @@ That is, if we stitched together pieces of the slope field, would we get a curve
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```{julia}
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#| hold: true
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#| echo: false
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#| cache: true
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## {{{euler_graph}}}
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function make_euler_graph(n)
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x, y = symbols("x, y")
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@@ -241,17 +240,7 @@ It is more work for the computer, but not for us, and clearly a much better appr
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## The Euler method
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```{julia}
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#| hold: true
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#| echo: false
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imgfile ="figures/euler.png"
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caption = """
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Figure from first publication of Euler's method. From [Gander and Wanner](http://www.unige.ch/~gander/Preprints/Ritz.pdf).
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"""
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ImageFile(:ODEs, imgfile, caption)
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```
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.](./figures/euler.png)
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The name of our function reflects the [mathematician](https://en.wikipedia.org/wiki/Leonhard_Euler) associated with the iteration:
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@@ -361,9 +350,13 @@ caption = """
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A child's bead game. What shape wire will produce the shortest time for a bed to slide from a top to the bottom?
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"""
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ImageFile(:ODEs, imgfile, caption)
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#ImageFile(:ODEs, imgfile, caption)
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nothing
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```
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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 $A$.
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@@ -378,16 +371,22 @@ caption = """
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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.
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From the [History of Math Archive](http://www-history.mcs.st-and.ac.uk/HistTopics/Brachistochrone.html).
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"""
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ImageFile(:ODEs, imgfile, caption)
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#ImageFile(:ODEs, imgfile, caption)
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nothing
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```
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.
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](./figures/galileo.png)
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This simulation also suggests that a curved path is better than the shorter straight one:
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```{julia}
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#| hold: true
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#| echo: false
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#| cache: true
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##{{{brach_graph}}}
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function brach(f, x0, vx0, y0, vy0, dt, n)
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@@ -603,13 +602,11 @@ $$
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We can try the Euler method here. A simple approach might be this iteration scheme:
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$$
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\begin{align*}
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x_{n+1} &= x_n + h,\\
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u_{n+1} &= u_n + h v_n,\\
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v_{n+1} &= v_n - h \cdot g/l \cdot \sin(u_n).
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\end{align*}
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$$
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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 $g=9.8$:
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