[Analysis] More examples

semester3/analysis-ii/cheat-sheet-rb/parts/05_diff.tex

@@ -100,10 +100,27 @@ $$
 
 \lemma \textbf{Chain Rule} $\quad g \circ f \text{ is differentiable on } X$
 \begin{align*}
-    & d(g \circ f)(x_0) &= dg\bigl( f(x_0) \bigr) \circ df(x_0) \\
-    & \textbf{J}_{g \circ f}(x_0) &= \textbf{J}_g\bigl( f(x_0) \bigr) \cdot \textbf{J}_f(x_0)
+    & d(f \circ g)(x_0) &= df\bigl( g(x_0) \bigr) \circ dg(x_0) \\
+    & \textbf{J}_{f \circ g}(x_0) &= \textbf{J}_f\bigl( g(x_0) \bigr) \cdot \textbf{J}_f(x_0)
 \end{align*}
-\subtext{$X \subset \R^n \text{ open},\quad Y \subset \R^m \text{ open},\quad f: X \to Y, g: Y \to \R^p, f,g \text{ diff.-able}$}
+\subtext{
+    $X \subset \R^n \text{ open},\quad Y \subset \R^m \text{ open},\quad f: X \to Y, g: Y \to \R^p, f,g \text{ diff.-able}$\\
+    $f \circ g$ meaning $f(g(x))$.
+}
+
+\begin{footnotesize}
+    \textbf{Example:} Let $\nabla f(- \frac{\sqrt{3}}{2}, \frac{3}{2}, 7) = (6, 2, 0)$.\\
+    We want $\partial_r f(\sqrt{3}, \frac{2}{3}\pi, 7)$ in cyl. coords, i.e. the 1st entry of $\textbf{J}_{f \circ g}$ with:
+    $$
+        g(r, \theta, z) = (r\cos(\theta), r\sin(\theta), z)
+    $$
+    Applying the Chain Rule:
+    $$
+        \textbf{J}_{f \circ g} = \textbf{J}_f\bigr( g(x_0) \bigl) \cdot \textbf{J}_g(x_0) = \underbrace{\textbf{J}_{f}\bigl(g(\sqrt{3}, \frac{3}{2}\pi, 7)\bigr)}_{(6, 2, 0)} \cdot \textbf{J}_g
+    $$
+    Where $\textbf{J}_g$ is the Jacobian for cylindrical coordinates.
+\end{footnotesize}
+
 
 \definition \textbf{Tangent Space}
 $$
@@ -111,6 +128,29 @@ $$
 $$
 \subtext{$X \subset \R^n \text{ open},\quad f: X \to \R^m \text{ diff.-able},\quad x_0 \in X,\quad u = df(x_0)$}
 
+\begin{footnotesize}
+    \textbf{Example:} Find $T_f\underbrace{(1, 1, 1)}_{x_0}$ for $S = \{\ \underbrace{(x, y, e^{x-y})}_{f(x,y)} \sep (x,y) \in \R^2 \} \subset \R^3$.
+    $$
+        u = df(x_0) = \textbf{J}_f(x_0) = 
+        \begin{bmatrix}
+            1       & 0 \\
+            0       & 1 \\
+            e^{x-y} & -e^{x-y}
+        \end{bmatrix}(x_0)
+        =
+        \begin{bmatrix}
+            1 & 0 \\
+            0 & 1 \\
+            1 & -1
+        \end{bmatrix}
+    $$
+    So we can define $T_f(x_0)$ as $g(x, y) = \underbrace{f(x_0)}_{(1, 1, 1)} + \textbf{J}_f(x_0) \cdot \begin{bmatrix}
+        x - x_{0, 1} \\
+        y - y_{0, 2}
+    \end{bmatrix}$.
+\end{footnotesize}
+
+
 \definition \textbf{Directional Derivative}
 $$
     D_v f(x_0) = \underset{t \neq 0 \to 0}{\lim} \frac{f(x_0 + tv) - f(x_0)}{t}
@@ -123,9 +163,11 @@ $$
 $$
 \subtext{$X \subset \R^n \text{ open},\quad f: X \to \R^m \text{ diff.-able},\quad v \neq 0 \in \R^n,\quad x_0 \in X$}
 
-\remark $D_vf$ is linear w.r.t $v$, so: $D_{v_1 + v_2}f = D_{v_1}f + D_{v_2}f$
+\begin{footnotesize}
+    \remark $D_vf$ is linear w.r.t $v$, so: $D_{v_1 + v_2}f = D_{v_1}f + D_{v_2}f$
 
-\remark $D_vf(x_0) = \nabla f(x_0) \cdot v = \big\| \nabla f(x_0) \big\| \cos(\theta)$\\
+    \remark $D_vf(x_0) = \nabla f(x_0) \cdot v = \big\| \nabla f(x_0) \big\| \cos(\theta)$\\
+\end{footnotesize}
 \subtext{In the case $f: X \to \R$, where $\theta$ is the angle between $v$ and $\nabla f(x_0)$}
 
 \newpage

semester3/analysis-ii/cheat-sheet-rb/parts/06_int.tex

@@ -304,4 +304,22 @@ $$
 $$
 \subtext{Same assumptions as above.}
 
-\remark The \textit{Gauss-Ostrogradski} Formula exists for $\R^3$.
+\footnotesize
+\remark The \textit{Gauss-Ostrogradski} Formula exists for $\R^3$.
+\normalsize
+
+\begin{footnotesize}
+    \textbf{Example:} Finding an Area using Green:
+    $$
+        \text{Area}(X) = \int_X 1\ dxdy
+    $$
+    To use Green, we want $f$ s.t. $(\partial_x f_y - \partial_y f_x) = 1$. Common Choices:
+    \begin{enumerate}
+        \item $f(x,y) = (0, x)$
+        \item $f(x,y) = (-y, 0)$
+    \end{enumerate}
+    We can then use Green, assuming we have a $\gamma$ for the Boundary of $X$:
+    $$
+        \text{Area}(X) = \int_X 1\ dxdy = \int_X \bigg(\partial_x f_y - \partial_y f_x \bigg) \ dxdy = \int_\gamma f(s)\ ds
+    $$
+\end{footnotesize}