align fix; theorem style; condition number

quarto/ODEs/differential_equations.qmd

@@ -71,12 +71,13 @@ $$
 
 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.
 
@@ -130,12 +131,13 @@ 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.
 
@@ -277,11 +279,12 @@ 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

@@ -602,12 +602,13 @@ $$
 
 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

@@ -86,12 +86,13 @@ $$
 
 Again, we can integrate to get an answer for any value $t$:
 
-
+$$
 \begin{align*}
 x(t) - x(t_0) &= \int_{t_0}^t \frac{dx}{dt} dt     \\
 &= (v_0t + \frac{1}{2}a t^2 - at_0 t) |_{t_0}^t    \\
 &= (v_0 - at_0)(t - t_0) + \frac{1}{2} a (t^2 - t_0^2).
 \end{align*}
+$$
 
 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:
 
@@ -336,11 +337,12 @@ 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:
 
@@ -667,12 +669,13 @@ 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.