[Analysis] Higher derivatives, change of variable, start taylor polynomial

semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/03_higher_diff.tex

@@ -1,2 +1,27 @@
 \newsectionNoPB
 \subsection{Higher derivatives}
+\shortdef $f$ is in class $C^1$ if $f$ is differentiable and all its partial derivatives are continuous.
+$f$ is of class $C^k$ if it is differentiable and each of its partial derivatives are in $C^{k - 1}$.
+If $f \in C^k(X; \R^m)$ for all $k \geq 1$, then $f \in C^\infty(X; \R^m)$\\
+% ────────────────────────────────────────────────────────────────────
+\setLabelNumber{all}{4}
+\compactproposition{Mixed derivatives commute} $\partial_{x, y} f = \partial_{y, x}$, as well as $\partial_{x, y, z} = \partial_{x, z, y} = \ldots$, etc (all mixed derivatives commute)
+Since we have symmetry, we can use the notation $\partial_{x_1^{m_1}, \ldots, x_n^{m_n}} f = \frac{\partial^k}{\partial x^m} f = D^m f = \partial^m f$,
+where $m = (m_1, \ldots, m_n)$ and $m_1 + \ldots + m_n = k$.
+There are ${n + k - 1 \choose k}$ possible values for $m$ and e.g. $(1, 1, 2)$ corresponds to the derivative $\frac{\partial^4 f}{\partial x \partial y \partial^2 z}$\\
+% ────────────────────────────────────────────────────────────────────
+\stepLabelNumber{all}
+\shortremark Due to linearity of the partial derivative $\partial_x^m(a f_1 + b f_2) = a \partial_x^m f_1 + b \partial_x^m f_2$\\
+% ────────────────────────────────────────────────────────────────────
+\stepLabelNumber{all}
+\compactex{Laplace operator} $f \in C^2(X)$, $\nabla f \in C_1(X; \R^n)$, so
+$\displaystyle \text{div}(\nabla f) = \sum_{i = 1}^{n} \frac{\partial}{\partial_{x_i}} \left( \frac{\partial f}{\partial_{x_i}} \right)
+    = \sum_{i = 1}^{n} \frac{\partial^2 f}{\partial x^2_i}$ (called \bi{Laplacian}, $\Delta f$)\\
+% ────────────────────────────────────────────────────────────────────
+\compactdef{Hessian} $f : X \rightarrow \R$ in $C^2$. For $x \in X$, the \bi{Hessian matrix} of $f$ at $x$ is the symmetric square matrix
+\vspace{-0.75pc}
+\begin{align*}
+    \text{Hess}_f(x) = (\partial_{x_i, x_j} f)_{1 \leq i, j \leq n} = H_f(x) \mediumhspace (\text{$i$-th row, $j$-th column})
+\end{align*}
+
+\drmvspace

semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/04_change_of_variable.tex

@@ -1,2 +1,5 @@
 \newsectionNoPB
 \subsection{Change of variable}
+The idea is to substitute variables for others that make the equation easier to solve.
+A common example is to switch to polar coordinates from cartesian coordinates, as already demonstrated with continuity checks
+% TODO: Add notes from TA notes

semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/05_taylor_polynomials.tex

@@ -1,2 +1,24 @@
 \newsectionNoPB
 \subsection{Taylor polynomials}
+\compactdef{Taylor polynomials}
+Let $f : X \rightarrow \R$ with $f \in C^k(X, \R)$ and $y \in X$. The Taylor-Polynomial of order $k$ of $f$ at $y$ is:
+\vspace{-0.75pc}
+\begin{align*}
+    T_k f(y; x - y) = \sum_{|i| \leq k} \frac{\partial_i f(y)(x - y)^i}{i!}
+\end{align*}
+
+\drmvspace\rmvspace
+% TODO: Find out what the \partial_1 notation means (likely TA notes 09)
+where $i$ is a \textit{multi-index}, so:
+\drmvspace
+\begin{multicols}{2}
+    \begin{itemize}[noitemsep]
+        \item $i = (i_1, \ldots, i_n)$ (each $i_j \geq 0$)
+        \item $|i| = i_1 + \ldots + i_n$, $\partial_i = \partial_1^{i_1} \ldots \partial_n^{i_n}$
+        \item $(x - y)^i = (x_1 - y_1)^{i_1} \cdot \ldots \cdot (x_n - y_n)^{i_n}$
+        \item $i! = i_1! \cdot \ldots \cdot i_n!$
+    \end{itemize}
+\end{multicols}
+
+\drmvspace\rmvspace
+The concept this formula uses is that we iterate through all possible partial derivatives of $f$ and assigns each a multi-index $i$