[Analysis] Cleaner structure

semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/00_continuity.tex

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+\stepcounter{subsection}
+\subsection{Continuity}
+\compactdef{Convergence in $\R^n$} Let $(x_k)_{k \in \N}$ where $x_k \in \R^n$ with $x_k = (x_{k, 1}, \ldots, x_{k, n})$ and let $y = (y_1, \ldots, y_n) \in \R^n$.
+$(x_k)$ converges to $y$ as $k \rightarrow +\infty$ if $\forall \varepsilon > 0 \smallhspace \exists N \geq 1$ s.t. $\forall n \geq N$ we have $||x_k - y|| < \varepsilon$\\
+% ────────────────────────────────────────────────────────────────────
+\shortlemma $(x_k)$ converges to $y$ as $k \rightarrow +\infty$ iff one of following equiv. statements holds:
+\bi{(1)} $\forall 1 \leq i \leq n$, the sequence $(x_{k, i})$ with $x_{k, i} \in \R$ converges to $y_i$
+\bi{(2)} $(||x_k - y||)$ converges to $0$ as $k \rightarrow +\infty$\\
+% ────────────────────────────────────────────────────────────────────
+\compactdef{Continuity} Let $X \subseteq \R^n$ and $f: X \rightarrow \R^m$.
+\bi{(1)} Let $x_0 \in X$. $f$ continuous in $\R^n$ if $\forall \varepsilon > 0 \smallhspace \exists \delta > 0$ s.t. if $x \in X$ satisfies $||x - x_0|| < \delta$,
+then $||f(x) - f(x_0)|| < \varepsilon$
+\bi{(2)} $f$ continuous \textit{on} $X$ if continuous at $x_0 \smallhspace \forall x_0 \in X$
+% TODO: Add tricks from TA slides here (week 05 / 04)
+% ────────────────────────────────────────────────────────────────────
+\shortproposition Let $X$ and $f$ as prev. Let $x_0 \in X$. $f$ continuous at $x_0$ iff $\forall (x_k)_{k \geq 1}$ in $X$ s.t.
+$x_k \rightarrow x_0$ as $k \rightarrow +\infty$, $(f(x_k))_{k \geq 1}$ in $\R^m$ converges to $f(x)$\\
+% ────────────────────────────────────────────────────────────────────
+\compactdef{Limit} Let $X$, $f$ and $x_0$ as prev. and $y \in \R^m$. $f$ \textit{has limit} $y$ as $x \rightarrow x_0$ with $x \neq x_0$ if
+$\forall \varepsilon > 0 \smallhspace \exists \delta > 0$ s.t. $\forall x \neq x_0 \in X, ||x - x_0|| < \delta$ we have $||f(x) - y|| < \varepsilon$.
+We write $\lim_{\elementstack{x \rightarrow x_0}{x \neq x_0}} f(x) = y$
+\shortremark Also possible without ass. that $x_0 \in X$
+% ────────────────────────────────────────────────────────────────────
+\shortproposition Let $X$, $f$, $x_0$ and $y$ as prev. We have $\lim_{\elementstack{x \rightarrow x_0}{x \neq x_0}} f(x) = y$
+iff $\forall (x_k)$ in $X$ s.t. $x_k \rightarrow x$ as $k \rightarrow +\infty$ and $x_k \neq x_0$ $(f(x_k))$ in $\R^m$ converges to $y$
+\stepLabelNumber{all}
+\shortproposition Let $X \subseteq \R^n$, $y \subseteq \R^m$, $p \in \N$ and let $f: X \rightarrow Y$ and $g: Y \rightarrow \R^p$ be cont. Then $g \circ f$ is continuous
+% ────────────────────────────────────────────────────────────────────
+\numberingOff
+\inlineremark To find the limits, we have two tricks (for $\limit{(x, y)}{(a, b)}$):
+\rmvspace
+\begin{enumerate}[noitemsep]
+    \item \bi{(Substitution)} Substitute $y = x + (b - a)$, then limit is $\limit{x}{(a - b)}$
+    \item \bi{(Polar coordinates)} Substitute $x = r \cos(\varphi)$ and $y = r \sin(\varphi)$ and the limit is $\limit{r}{0}$
+\end{enumerate}
+
+\numberingOn\rmvspace
+\shortex \bi{(1)} $f_1 : \R^n \rightarrow \R^{m_1}$ and $f_2 : \R^n \rightarrow \R^{m_2}$ continuous
+$\Rightarrow f = (f_1, f_2): \R^n \rightarrow \R^{m_1 + m_2}$ is continuous (Cartesian product)
+\bi{(2)} Any linear map $f: \R^n \rightarrow \R^m$ is continuous. In particular, the identity map is continuous
+\bi{(3)} If $f_1, \ldots, f_n$ continuous, then $f(x_1, \ldots, x_n) = f_1(x_1) \cdot \ldots \cdot f_n(x_n)$ is continuous
+\bi{(4)} Polynomials in $x_1, \ldots, x_n$ are continuous
+\bi{(5)} $f_1f_2$ is continuous if $f_1$ and $f_2$ are continuous and if $f_2(x) \neq 0 \smallhspace \forall x \in X$, then $f_1 \div f_2$ is continuous.
+(see Theorem 2.1.8 in Analysis I)\\
+\bi{(6)} If both $f$ and $g$ have limits, then $\displaystyle \limit{x}{x_0}(f(x) + g(x)) = \limit{x}{x_0} f(x) + \limit{x}{x_0} g(x)$ and analogous for $\times$
+\bi{(7)} If $f: \R^2 \rightarrow \R$ continuous, then $g(x) = f(x, y_0)$ for $y_0 \in \R$ is continuous. The converse is not true\\
+% ────────────────────────────────────────────────────────────────────
+\shortdef \bi{(1)} $X \subseteq \R^n$ is \bi{bounded} if the set of $||x||$ for $x \in X$ is bounded in $\R$
+\bi{(2)} $X \subseteq \R^n$ is \bi{closed} if $\forall (x_k)$ in $X$ that converge in $\R^n$ to some vector $y \in \R^n$, we have $y \in X$
+% Closed simple explanation: "border" included or not (and is not in case of open disc because limit is outside, i.e. limit of a_n = (n / (n + 1), 0) is 1)
+\bi{(3)} $X \subseteq \R^n$ is \bi{compact} if it is bounded and closed\\
+% ────────────────────────────────────────────────────────────────────
+\shortex \bi{(1)} $\emptyset$ and $\R^n$ are closed.
+\bi{(2)} The \textit{open} disc $D = \{ x \in \R^n : ||x - x_0|| < r \}$ for $r > 0$ and $x_0 \in \R^n$ is bounded and not closed.
+\bi{(3)} The \textit{closed} disc $\Delta = \{ x \in \R^n : ||x - x_0|| \leq r \}$ is bounded and closed. In particular, a closed interval is a closed set.
+An interval is compact if it is bounded
+\bi{(4)} If $X_1 \subseteq \R^n$ and $X_2 \subseteq \R^m$ are bounded (also closed or compact), then so is $X_1 \times X_2 \subseteq \R^{n + m}$\\
+% ────────────────────────────────────────────────────────────────────
+\shortproposition Let $f: \R^n \rightarrow \R^m$ be a continuous map. For any closed $Y \subseteq \R^m$,
+the set $f^{-1}(Y) = \{ x \in \R^n : f(x) \in Y \} \subseteq \R^n$ is closed\\
+% ────────────────────────────────────────────────────────────────────
+\shortex The \bi{zero set} $Z = \{ x\in \R^n : f(x) = 0 \}$ is closed in $\R^n$ because $\{ 0 \} \subseteq \R$ is closed.
+More generally: for any $r \geq 0$, $\{ x \in \R^n : |f(x)| \leq r \}$ is $f^{-1}([-r, r])$ and is closed, since $[-r, r]$ is closed.
+Furthermore: $\{ x\in \R^3 : ||x - x_0|| = r \}$ is closed\\
+% ────────────────────────────────────────────────────────────────────
+\shorttheorem Let $(X \neq \emptyset) \subseteq \R^n$ compact and $f: X \rightarrow \R$ continuous. 
+Then $f$ bounded, has $\max$ and $\min$, i.e. $\exists x_+, x_- \in X$ s.t. $\displaystyle f(x_+) = \sup_{x \in X} f(x)$ and $\displaystyle f(x_-) = \inf_{x \in X} f(x)$
+% ────────────────────────────────────────────────────────────────────

semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/01_partial_derivatives.tex

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+\newsectionNoPB
+\subsection{Partial derivatives}
+\shortdef $X \subseteq \R^n$ \bi{open} if for any $x = (x_1, \ldots, x_n) \in X$ $\exists \delta > 0$ s.t.
+$\{ y = (y_1, \ldots, y_n) \in \R^n : |x_i - y_i| < \delta \smallhspace \forall i \}$ is contained in $X$.
+(= changing a coordinate of $x$ by $< \delta \rightarrow x' \in X$)
+\shortproposition $X \subseteq \R^n$ open $\Leftrightarrow$ \bi{complement} $Y = \{ x \in \R^n : x \notin X \}$ is closed
+\shortcorollary If $f: \R^n \rightarrow \R^m$ cont. and $Y \subseteq \R^m$ open, then $f^{-1}(Y)$ is open in $\R^n$
+\shortex \bi{(1)} $\emptyset$ and $\R^n$ are both open and closed.
+\bi{(2)} Open ball $D = \{ x \in \R^n : ||x - x_0|| < r \}$ is open in $\R^n$ ($x_0$ the center and $r$ radius)
+\bi{(3)} $I_1 \times \dots \times I_n$ is open in $\R^n$ for $I_i$ open
+\bi{(4)} $X \subseteq \R^n$ open $\Leftrightarrow$ $\forall x \in X \exists \delta > 0$ s.t. open ball of center $x$ and radius $\delta$ is contained in $X$
+
+\compactdef{Partial derivative} Let $X \subseteq \R^n$ open, $f: X \rightarrow \R^m$ and $1 \leq i \leq n$.
+Then $f$ has partial derivative on $X$ with respect to the $i$-th variable (or coordinate),
+if $\forall x_0 = (x_{0, 1}, \ldots, x_{0, n}) \in X$, $g(t) = f(x_{0, 1}, \ldots, x_{0, i - 1}, t, x_{0, i + 1}, x_{0, n})$ on set
+$I = \{ t \in \R : (x_{0, 1}, \ldots, x_{0, i - 1}, t, x_{0, i + 1}, \ldots, x_{0, n}) \in X \}$ is differentiable at $t = x_{0, i}$.
+The derivative $g'(x_{0, i})$ at $x_{0, i}$ is denoted:
+$\frac{\partial f}{\partial x_i}(x_0), \partial_{x_i} f(x_0) \text{ or } \partial_i f(x_0)$\\
+%
+\shortproposition Let $X \subseteq \R^n$ open, $f, g : X \rightarrow \R^m$ and $1 \leq i \leq n$. Then:
+\bi{(1)} If $f$ \& $g$ have $\partial_i$ on $X$, then so does $f + g$ and $\partial_{x_i} (f + g) = \partial_{x_i}(f) + \partial_{x_i}(g)$
+\bi{(2)} If $m = 1$ (i.e. $\R^1$) and $f$ \& $g$ have $\partial_i$ on $X$, then so does $fg$ and $\partial_{x_i} (fg) = \partial_{x_i}(f)g + f \partial_{x_i}(g)$
+and if $g(x) \neq 0 \smallhspace \forall x \in X$, then if $f \div g$ has $\partial_i$ on $X$, then so does $f \div g$ and
+$\partial_{x_i}(f \div g) = (\partial_{x_i}(f) g - f \partial_{x_i}(g)) \div g^2$