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[Analysis] Better structure

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Janis Hutz
2025-10-24 11:06:17 +02:00
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semester3/analysis-ii/analysis-ii-cheat-sheet.pdf
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semester3/analysis-ii/parts/vectors/differentiation/00_continuity.tex
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\stepcounter{subsection} \stepcounter{subsection}
\subsection{Continuity} \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 $y = (y_1, \ldots, y_n) \in \R^n$. \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 $||x_k - y|| < \varepsilon$ $(x_k)$ 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: \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{(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 $k \rightarrow +\infty$ \bi{(2)} $(||x_k - y||)$ converges to $0$ as $k \rightarrow +\infty$\\
% ──────────────────────────────────────────────────────────────────── % ────────────────────────────────────────────────────────────────────
\compactdef{Continuity} Let $X \subseteq \R^n$ and $f: X \rightarrow \R^m$. \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 $||x - x_0|| < \delta$, \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 $||f(x) - f(x_0)|| < \varepsilon$ then $||f(x) - f(x_0)|| < \varepsilon$
\bi{(2)} $f$ continuous \textit{on} $X$ if continuous at $x_0 \smallhspace \forall x_0 \in X$ \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. \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 $f(x)$\\ $x_k \rightarrow x_0$ as $k \rightarrow +\infty$, $(f(x_k))_{k \geq 1}$ in $\R^m$ converges to $f(x)$\\