align fix; theorem style; condition number

quarto/differentiable_vector_calculus/vector_fields.qmd

@@ -191,11 +191,12 @@ surface(unzip(Phi.(thetas, phis'))...)
 The partial derivatives of each component, $\partial{\Phi}/\partial{\theta}$ and $\partial{\Phi}/\partial{\phi}$, can be computed directly:
 
 
-
+$$
 \begin{align*}
 \partial{\Phi}/\partial{\theta} &= \langle -\sin(\phi)\sin(\theta), \sin(\phi)\cos(\theta),0 \rangle,\\
 \partial{\Phi}/\partial{\phi} &= \langle \cos(\phi)\cos(\theta), \cos(\phi)\sin(\theta), -\sin(\phi) \rangle.
 \end{align*}
+$$
 
 
 Using `SymPy`, we can compute through:
@@ -359,7 +360,7 @@ where $\epsilon(h) \rightarrow \vec{0}$ as $h \rightarrow \vec{0}$.
 We have, using this for *both* $F$ and $G$:
 
 
-
+$$
 \begin{align*}
 F(G(a + \vec{h})) - F(G(a)) &=
 F(G(a) + (dG_a \cdot \vec{h} + \epsilon_G \vec{h})) - F(G(a))\\
@@ -367,18 +368,20 @@ F(G(a) + (dG_a \cdot \vec{h} + \epsilon_G \vec{h})) - F(G(a))\\
 &+ \quad\epsilon_F (dG_a \cdot \vec{h} + \epsilon_G \vec{h}) - F(G(a))\\
 &= dF_{G(a)} \cdot (dG_a \cdot \vec{h})  +  dF_{G(a)} \cdot (\epsilon_G \vec{h}) + \epsilon_F (dG_a \cdot \vec{h}) + (\epsilon_F \cdot \epsilon_G\vec{h})
 \end{align*}
+$$
 
 
 The last line uses the linearity of $dF$ to isolate $dF_{G(a)} \cdot (dG_a \cdot \vec{h})$. Factoring out $\vec{h}$ and taking norms gives:
 
 
-
+$$
 \begin{align*}
 \frac{\| F(G(a+\vec{h})) - F(G(a)) - dF_{G(a)}dG_a \cdot \vec{h} \|}{\| \vec{h} \|} &=
 \frac{\|  dF_{G(a)}\cdot(\epsilon_G\vec{h}) + \epsilon_F (dG_a\cdot \vec{h}) + (\epsilon_F\cdot\epsilon_G\vec{h}) \|}{\| \vec{h} \|} \\
 &\leq \|  dF_{G(a)}\cdot\epsilon_G + \epsilon_F (dG_a) + \epsilon_F\cdot\epsilon_G \|\frac{\|\vec{h}\|}{\| \vec{h} \|}\\
 &\rightarrow 0.
 \end{align*}
+$$
 
 
 ### Examples
@@ -660,7 +663,7 @@ det(A1), 1/det(A2)
 The technique of *implicit differentiation* is a useful one, as it allows derivatives of more complicated expressions to be found. The main idea, expressed here with three variables is if an equation may be viewed as $F(x,y,z) = c$, $c$ a constant, then $z=\phi(x,y)$ may be viewed as a function of $x$ and $y$. Hence, we can use the chain rule to find: $\partial z / \partial x$ and $\partial z /\partial y$. Let $G(x,y) = \langle x, y, \phi(x,y) \rangle$ and then differentiation $(F \circ G)(x,y) = c$:
 
 
-
+$$
 \begin{align*}
 0 &= dF_{G(x,y)} \circ dG_{\langle x, y\rangle}\\
 &= [\frac{\partial F}{\partial x}\quad \frac{\partial F}{\partial y}\quad \frac{\partial F}{\partial z}](G(x,y)) \cdot
@@ -670,6 +673,7 @@ The technique of *implicit differentiation* is a useful one, as it allows deriva
 \frac{\partial \phi}{\partial x} & \frac{\partial \phi}{\partial y}
 \end{bmatrix}.
 \end{align*}
+$$
 
 
 Solving yields
@@ -685,14 +689,17 @@ Where the right hand side of each is evaluated at $G(x,y)$.
 
 When can it be reasonably assumed that such a function $z= \phi(x,y)$ exists?
 
-
-The [Implicit Function Theorem](https://en.wikipedia.org/wiki/Implicit_function_theorem) provides a statement (slightly abridged here):
+::: {.callout-note icon=false}
+The [Implicit Function Theorem](https://en.wikipedia.org/wiki/Implicit_function_theorem) (slightly abridged)
 
 
-> Let $F:R^{n+m} \rightarrow R^m$ be a continuously differentiable function and let $R^{n+m}$ have (compactly defined) coordinates $\langle \vec{x}, \vec{y} \rangle$, Fix a point $\langle \vec{a}, \vec{b} \rangle$ with $F(\vec{a}, \vec{b}) = \vec{0}$. Let $J_{F, \vec{y}}(\vec{a}, \vec{b})$ be the Jacobian restricted to *just* the $y$ variables. ($J$ is $m \times m$.) If this matrix has non-zero determinant (it is invertible), then there exists an open set $U$ containing $\vec{a}$ and a *unique* continuously differentiable function $G: U \subset R^n \rightarrow R^m$ such that $G(\vec{a}) = \vec{b}$, $F(\vec{x}, G(\vec{x})) = 0$ for $\vec x$ in $U$. Moreover, the partial derivatives of $G$ are given by the matrix product:
->
-> $\frac{\partial G}{\partial x_j}(\vec{x}) = - [J_{F, \vec{y}}(x, F(\vec{x}))]^{-1} \left[\frac{\partial F}{\partial x_j}(x, G(\vec{x}))\right].$
+Let $F:R^{n+m} \rightarrow R^m$ be a continuously differentiable function and let $R^{n+m}$ have (compactly defined) coordinates $\langle \vec{x}, \vec{y} \rangle$, Fix a point $\langle \vec{a}, \vec{b} \rangle$ with $F(\vec{a}, \vec{b}) = \vec{0}$. Let $J_{F, \vec{y}}(\vec{a}, \vec{b})$ be the Jacobian restricted to *just* the $y$ variables. ($J$ is $m \times m$.) If this matrix has non-zero determinant (it is invertible), then there exists an open set $U$ containing $\vec{a}$ and a *unique* continuously differentiable function $G: U \subset R^n \rightarrow R^m$ such that $G(\vec{a}) = \vec{b}$, $F(\vec{x}, G(\vec{x})) = 0$ for $\vec x$ in $U$. Moreover, the partial derivatives of $G$ are given by the matrix product:
 
+$$
+\frac{\partial G}{\partial x_j}(\vec{x}) = - [J_{F, \vec{y}}(x, F(\vec{x}))]^{-1} \left[\frac{\partial F}{\partial x_j}(x, G(\vec{x}))\right].
+$$
+
+:::
 
 
 ---