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Diffstat (limited to '')
| -rw-r--r-- | vorlesungen/slides/10/ableitung-exp.tex | 60 | ||||
| -rw-r--r-- | vorlesungen/slides/10/intro.tex | 45 | ||||
| -rw-r--r-- | vorlesungen/slides/10/matrix-dgl.tex (renamed from vorlesungen/slides/10/matrix-vektor-dgl.tex) | 2 | ||||
| -rw-r--r-- | vorlesungen/slides/10/n-zu-1.tex | 98 | ||||
| -rw-r--r-- | vorlesungen/slides/10/potenzreihenmethode.tex | 91 | ||||
| -rw-r--r-- | vorlesungen/slides/10/repetition.tex | 119 | ||||
| -rw-r--r-- | vorlesungen/slides/10/so2.tex | 243 | ||||
| -rw-r--r-- | vorlesungen/slides/10/taylor.tex | 194 | ||||
| -rw-r--r-- | vorlesungen/slides/10/vektorfelder.mp | 241 | ||||
| -rw-r--r-- | vorlesungen/slides/10/vektorfelder.tex | 82 |
10 files changed, 848 insertions, 327 deletions
diff --git a/vorlesungen/slides/10/ableitung-exp.tex b/vorlesungen/slides/10/ableitung-exp.tex new file mode 100644 index 0000000..10ce191 --- /dev/null +++ b/vorlesungen/slides/10/ableitung-exp.tex @@ -0,0 +1,60 @@ +% +% ableitung-exp.tex -- Ableitung von exp(x) +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% Erstellt durch Roy Seitz +% +% !TeX spellcheck = de_CH +\bgroup +\begin{frame}[t] + \setlength{\abovedisplayskip}{5pt} + \setlength{\belowdisplayskip}{5pt} + %\frametitle{Ableitung von $\exp(x)$} + %\vspace{-20pt} + \begin{columns}[t,onlytextwidth] + \begin{column}{0.48\textwidth} + \begin{block}{Ableitung von $\exp(at)$} + \begin{align*} + \frac{d}{dt} \exp(at) + &= + \frac{d}{dt} \sum_{k=0}^{\infty} a^k \frac{t^k}{k!} + \\ + &\uncover<2->{ + = \sum_{k=0}^{\infty} a^k\frac{kt^{k-1}}{k(k-1)!} + } + \\ + &\uncover<3->{ + = a \sum_{k=1}^{\infty} + a^{k-1}\frac{t^{k-1}}{(k-1)!} + } + \\ + &\uncover<4->{ + = a \exp(at) + } + \end{align*} + \end{block} + \end{column} + \begin{column}{0.48\textwidth} + \uncover<5->{ + \begin{block}{Ableitung von $\exp(At)$} + \begin{align*} + \frac{d}{dt} \exp(At) + &= + \frac{d}{dt} \sum_{k=0}^{\infty} A^k \frac{t^k}{k!} + \\ + &= + \sum_{k=0}^{\infty} A^k\frac{kt^{k-1}}{k(k-1)!} + \\ + &= + A \sum_{k=1}^{\infty} A^{k-1}\frac{t^{k-1}}{(k-1)!} + \\ + &= + A \exp(At) + \end{align*} + \end{block} + } + \end{column} + \end{columns} +\end{frame} + +\egroup diff --git a/vorlesungen/slides/10/intro.tex b/vorlesungen/slides/10/intro.tex new file mode 100644 index 0000000..276bf49 --- /dev/null +++ b/vorlesungen/slides/10/intro.tex @@ -0,0 +1,45 @@ +% +% intro.tex -- Repetition Lie-Gruppen und -Algebren +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% Erstellt durch Roy Seitz +% +% !TeX spellcheck = de_CH +\bgroup + + + +\begin{frame}[t] + \setlength{\abovedisplayskip}{5pt} + \setlength{\belowdisplayskip}{5pt} +% \frametitle{Repetition} +% \vspace{-20pt} + \begin{block}{Offene Fragen} + \begin{itemize}[<+->] + \item Woher kommt die Exponentialfunktion? + \begin{fleqn} + \[ + \exp(At) + = + 1 + + At + + A^2\frac{t^2}{2} + + A^3\frac{t^3}{3!} + + \ldots + \] + \end{fleqn} + \item Wie löst man eine Matrix-DGL? + \begin{fleqn} + \[ + \dot\gamma(t) = A\gamma(t), + \qquad + \gamma(t) \in G \subset M_n + \] + \end{fleqn} + \item Lie-Gruppen und Lie-Algebren + \item Was bedeutet $\exp(At)$? + \end{itemize} + \end{block} +\end{frame} + +\egroup diff --git a/vorlesungen/slides/10/matrix-vektor-dgl.tex b/vorlesungen/slides/10/matrix-dgl.tex index f7bd995..ae68fb1 100644 --- a/vorlesungen/slides/10/matrix-vektor-dgl.tex +++ b/vorlesungen/slides/10/matrix-dgl.tex @@ -1,5 +1,5 @@ % -% matrix-vektor-dgl.tex -- DGL mit Matrix-Koeffizienten und Vektor-Variablen +% matrix-dgl.tex -- Matrix-Differentialgleichungen % % (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule % Erstellt durch Roy Seitz diff --git a/vorlesungen/slides/10/n-zu-1.tex b/vorlesungen/slides/10/n-zu-1.tex index 737df03..09475ad 100644 --- a/vorlesungen/slides/10/n-zu-1.tex +++ b/vorlesungen/slides/10/n-zu-1.tex @@ -7,51 +7,57 @@ % !TeX spellcheck = de_CH \bgroup \begin{frame}[t] -\setlength{\abovedisplayskip}{5pt} -\setlength{\belowdisplayskip}{5pt} -\frametitle{Reicht $1.$ Ordnung?} -\vspace{-20pt} -\begin{columns}[t,onlytextwidth] -\begin{column}{0.48\textwidth} -\begin{block}{Beispiel: DGL 3.~Ordnung} \vspace*{-1ex} - \begin{align*} - x^{(3)} + a_2 \ddot x + a_1 \dot x + a_0 x = 0 \\ - \Rightarrow - x^{(3)} = -a_2 \ddot x - a_1 \dot x - a_0 x - \end{align*} -\end{block} -\begin{block}{Ziel: Nur noch 1.~Ableitungen} - Einführen neuer Variablen: - \begin{align*} - x_0 &\coloneqq x & - x_1 &\coloneqq \dot x & - x_2 &\coloneqq \ddot x - \end{align*} -System von Gleichungen 1.~Ordnung - \begin{align*} - \dot x_0 &= x_1 \\ - \dot x_1 &= x_2 \\ - \dot x_2 &= -a_2 x_2 - a_1 x_1 - a_0 x_0 -\end{align*} -\end{block} -\end{column} -\begin{column}{0.48\textwidth} -\begin{block}{Als Vektor-Gleichung} \vspace*{-1ex} - \begin{align*} - \frac{d}{dt} - \begin{pmatrix} x_0 \\ x_1 \\ x_2 \end{pmatrix} - = \begin{pmatrix} - 0 & 1 & 0 \\ - 0 & 0 & 1 \\ - -a_0 & -a_1 & -a_2 - \end{pmatrix} - \begin{pmatrix} x_0 \\ x_1 \\ x_2 \end{pmatrix} - \end{align*} - - Geht für jede lineare Differentialgleichung! - -\end{block} -\end{column} -\end{columns} + \setlength{\abovedisplayskip}{5pt} + \setlength{\belowdisplayskip}{5pt} + %\frametitle{Reicht $1.$ Ordnung?} + %\vspace{-20pt} + \begin{columns}[t,onlytextwidth] + \begin{column}{0.48\textwidth} + \uncover<1->{ + \begin{block}{Beispiel: DGL 3.~Ordnung} \vspace*{-1ex} + \begin{align*} + x^{(3)} + a_2 \ddot x + a_1 \dot x + a_0 x = 0 \\ + \Rightarrow + x^{(3)} = -a_2 \ddot x - a_1 \dot x - a_0 x + \end{align*} + \end{block} + } + \uncover<2->{ + \begin{block}{Ziel: Nur noch 1.~Ableitungen} + Einführen neuer Variablen: + \begin{align*} + x_0 &\coloneqq x & + x_1 &\coloneqq \dot x & + x_2 &\coloneqq \ddot x + \end{align*} + System von Gleichungen 1.~Ordnung + \begin{align*} + \dot x_0 &= x_1 \\ + \dot x_1 &= x_2 \\ + \dot x_2 &= -a_2 x_2 - a_1 x_1 - a_0 x_0 + \end{align*} + \end{block} + } + \end{column} + \uncover<3->{ + \begin{column}{0.48\textwidth} + \begin{block}{Als Vektor-Gleichung} \vspace*{-1ex} + \begin{align*} + \frac{d}{dt} + \begin{pmatrix} x_0 \\ x_1 \\ x_2 \end{pmatrix} + = \begin{pmatrix} + 0 & 1 & 0 \\ + 0 & 0 & 1 \\ + -a_0 & -a_1 & -a_2 + \end{pmatrix} + \begin{pmatrix} x_0 \\ x_1 \\ x_2 \end{pmatrix} + \end{align*} + + \uncover<4->{Geht für jede lineare Differentialgleichung!} + + \end{block} + \end{column} + } + \end{columns} \end{frame} \egroup diff --git a/vorlesungen/slides/10/potenzreihenmethode.tex b/vorlesungen/slides/10/potenzreihenmethode.tex new file mode 100644 index 0000000..1715134 --- /dev/null +++ b/vorlesungen/slides/10/potenzreihenmethode.tex @@ -0,0 +1,91 @@ +% +% potenzreihenmethode.tex +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% Bearbeitet durch Roy Seitz +% +\begin{frame}[t] +\setlength{\abovedisplayskip}{5pt} +\setlength{\belowdisplayskip}{5pt} +\frametitle{Potenzreihenmethode} +\vspace{-15pt} +\begin{columns}[t,onlytextwidth] +\begin{column}{0.48\textwidth} +\begin{block}{Lineare Differentialgleichung} +\begin{align*} +x'&=ax&&\Rightarrow&x'-ax&=0 +\\ +x(0)&=C +\end{align*} +\end{block} +\end{column} +\begin{column}{0.48\textwidth} +\uncover<2->{% +\begin{block}{Potenzreihenansatz} +\begin{align*} +x(t) +&= +a_0+ a_1t + a_2t^2 + \dots +\\ +x(0)&=a_0=C +\end{align*} +\end{block}} +\end{column} +\end{columns} +\uncover<3->{% +\begin{block}{Lösung} +\[ +\arraycolsep=1.4pt +\begin{array}{rcrcrcrcrcr} +\uncover<3->{ x'(t)} + \uncover<5->{ + &=&\phantom{(} a_1\phantom{\mathstrut-aa_0)} + &+& 2a_2\phantom{\mathstrut-aa_1)}t + &+& 3a_3\phantom{\mathstrut-aa_2)}t^2 + &+& 4a_4\phantom{\mathstrut-aa_3)}t^3 + &+& \dots}\\ +\uncover<3->{-ax(t)} + \uncover<6->{ + &=&\mathstrut-aa_0 \phantom{)} + &-& aa_1\phantom{)}t + &-& aa_2\phantom{)}t^2 + &-& aa_3\phantom{)}t^3 + &-& \dots}\\[2pt] +\hline +\\[-10pt] +\uncover<3->{0} + \uncover<7->{ + &=&(a_1-aa_0) + &+& (2a_2-aa_1)t + &+& (3a_3-aa_2)t^2 + &+& (4a_4-aa_3)t^3 + &+& \dots}\\ +\end{array} +\] +\begin{align*} +\uncover<4->{ +a_0&=C}\uncover<8->{, +\quad +a_1=aa_0=aC}\uncover<9->{, +\quad +a_2=\frac12a^2C}\uncover<10->{, +\quad +a_3=\frac16a^3C}\uncover<11->{, +\ldots, +a_k=\frac1{k!}a^kC} +\hspace{3cm} +\\ +\uncover<4->{ +\Rightarrow x(t) &= C}\uncover<8->{+Cat}\uncover<9->{ + C\frac12(at)^2} +\uncover<10->{ + C \frac16(at)^3} +\uncover<11->{ + \dots+C\frac{1}{k!}(at)^k+\dots} +\ifthenelse{\boolean{presentation}}{ +\only<12>{ += +C\sum_{k=0}^\infty \frac{(at)^k}{k!}} +}{} +\uncover<13->{= +C\exp(at)} +\end{align*} +\end{block}} +\end{frame} diff --git a/vorlesungen/slides/10/repetition.tex b/vorlesungen/slides/10/repetition.tex new file mode 100644 index 0000000..7c007ca --- /dev/null +++ b/vorlesungen/slides/10/repetition.tex @@ -0,0 +1,119 @@ +% +% repetition.tex -- Repetition Lie-Gruppen und -Algebren +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% Erstellt durch Roy Seitz +% +% !TeX spellcheck = de_CH +\bgroup + +\begin{frame}[t] + \setlength{\abovedisplayskip}{5pt} + \setlength{\belowdisplayskip}{5pt} + \frametitle{Repetition} + \vspace{-20pt} + \begin{columns}[t,onlytextwidth] + \begin{column}{0.48\textwidth} + \uncover<1->{ + \begin{block}{Lie-Gruppe} + Kontinuierliche Matrix-Gruppe $G$ mit bestimmter Eigenschaft + \end{block} + } + \uncover<3->{ + \begin{block}{Ein-Parameter-Untergruppe} + Darstellung der Lie-Gruppe $G$: + \[ + \gamma \colon \mathbb R \to G + : \quad + t \mapsto \gamma(t), + \] + so dass + \[ \gamma(s + t) = \gamma(t) \gamma(s). \] + \end{block} + } + \end{column} + \begin{column}{0.48\textwidth} + \uncover<2->{ + \begin{block}{Beispiel} + Volumen-erhaltende Abbildungen: + \[ \gSL2R= \{A \in M_2 \,|\, \det(A) = 1\} .\] + \begin{align*} + \uncover<4->{ \gamma_x(t) } + & + \uncover<4->{= \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix} } + \\ + \uncover<5->{ \gamma_y(t) } + & + \uncover<5->{= \begin{pmatrix} 1 & 0 \\ t & 1 \end{pmatrix} } + \\ + \uncover<6->{ \gamma_h(t)} + & + \uncover<6->{= \begin{pmatrix} e^t & 0 \\ 0 & e^{-t} \end{pmatrix} } + \end{align*} + \end{block} + } + \end{column} + \end{columns} +\end{frame} + + +\begin{frame}[t] + \setlength{\abovedisplayskip}{5pt} + \setlength{\belowdisplayskip}{5pt} + \frametitle{Repetition} + \vspace{-20pt} + \begin{columns}[t,onlytextwidth] + \begin{column}{0.48\textwidth} + \uncover<1->{ + \begin{block}{Lie-Algebra aus Lie-Gruppe} + Ableitungen der Ein-Parameter-Untergruppen: + \begin{align*} + G &\to \mathcal A \\ + \gamma &\mapsto \dot\gamma(0) + \end{align*} + \uncover<3->{ + Lie-Klammer als Produkt: + \[ [A, B] = AB - BA \in \mathcal A \] + } + \end{block} + } + \uncover<7->{\vspace*{-4ex} + \begin{block}{Lie-Gruppe aus Lie-Algebra} + Lösung der Differentialgleichung: + \[ + \dot\gamma(t) = A\gamma(t) + \quad \text{mit} \quad + A = \dot\gamma(0) + \] + \[ + \Rightarrow \gamma(t) = \exp(At) + \] + \end{block} + } + \end{column} + \begin{column}{0.48\textwidth} + \uncover<2->{ + \begin{block}{Beispiel} + Lie-Algebra von \gSL2R: + \[ \asl2R = \{ A \in M_2 \,|\, \Spur(A) = 0 \} \] + \end{block} + } + \begin{align*} + \uncover<4->{ X(t) } + & + \uncover<4->{= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} } + \\ + \uncover<5->{ Y(t) } + & + \uncover<5->{= \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} } + \\ + \uncover<6->{ H(t) } + & + \uncover<6->{= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} } + \end{align*} + + \end{column} + \end{columns} +\end{frame} + +\egroup diff --git a/vorlesungen/slides/10/so2.tex b/vorlesungen/slides/10/so2.tex index e3f74ae..dcbcdc8 100644 --- a/vorlesungen/slides/10/so2.tex +++ b/vorlesungen/slides/10/so2.tex @@ -7,132 +7,135 @@ % !TeX spellcheck = de_CH \bgroup -\newcommand{\gSL}[2]{\ensuremath{\text{SL}(#1, \mathbb{#2})}} -\newcommand{\gSO}[1]{\ensuremath{\text{SO}(#1)}} -\newcommand{\gGL}[2]{\ensuremath{\text{GL}(#1, \mathbb #2)}} +\begin{frame}[t] + \setlength{\abovedisplayskip}{5pt} + \setlength{\belowdisplayskip}{5pt} + \frametitle{Von der Lie-Gruppe zur -Algebra} + \vspace{-20pt} + \begin{columns}[t,onlytextwidth] + \begin{column}{0.48\textwidth} + \uncover<1->{ + \begin{block}{Lie-Gruppe} + Darstellung von \gSO2: + \begin{align*} + \mathbb R + &\to + \gSO2 + \\ + t + &\mapsto + \begin{pmatrix} + \cos t & -\sin t \\ + \sin t & \phantom-\cos t + \end{pmatrix} + \end{align*} + \end{block} + } + \uncover<2->{ + \begin{block}{Ableitung am neutralen Element} + \begin{align*} + \frac{d}{d t} + & + \left. + \begin{pmatrix} + \cos t & -\sin t \\ + \sin t & \phantom-\cos t + \end{pmatrix} + \right|_{ t = 0} + \\ + = + & + \begin{pmatrix} -\sin0 & -\cos0 \\ \phantom-\cos0 & -\sin0 \end{pmatrix} + = + \begin{pmatrix} 0 & -1 \\ 1 & \phantom-0 \end{pmatrix} + \end{align*} + \end{block} + } + \end{column} + \begin{column}{0.48\textwidth} + \uncover<3->{ + \begin{block}{Lie-Algebra} + Darstellung von \aso2: + \begin{align*} + \mathbb R + &\to + \aso2 + \\ + t + &\mapsto + \begin{pmatrix} + 0 & -t \\ + t & \phantom-0 + \end{pmatrix} + \end{align*} + \end{block} + } + \end{column} + \end{columns} +\end{frame} -\newcommand{\asl}[2]{\ensuremath{\mathfrak{sl}(#1, \mathbb{#2})}} -\newcommand{\aso}[1]{\ensuremath{\mathfrak{so}(#1)}} -\newcommand{\agl}[2]{\ensuremath{\mathfrak{gl}(#1, \mathbb #2)}} \begin{frame}[t] -\setlength{\abovedisplayskip}{5pt} -\setlength{\belowdisplayskip}{5pt} -\frametitle{Von der Lie-Gruppe zur -Algebra} -\vspace{-20pt} -\begin{columns}[t,onlytextwidth] -\begin{column}{0.48\textwidth} - \begin{block}{Lie-Gruppe} - Darstellung von \gSO2: - \begin{align*} - \mathbb R - &\to - \gSO2 - \\ - t - &\mapsto - \begin{pmatrix} - \cos t & -\sin t \\ - \sin t & \phantom-\cos t - \end{pmatrix} - \end{align*} - \end{block} - \begin{block}{Ableitung am neutralen Element} - \begin{align*} - \frac{d}{d t} - & - \left. + \setlength{\abovedisplayskip}{5pt} + \setlength{\belowdisplayskip}{5pt} + \frametitle{Von der Lie-Algebra zur -Gruppe} + \vspace{-20pt} + \begin{columns}[t,onlytextwidth] + \begin{column}{0.48\textwidth} + \uncover<1->{ + \begin{block}{Differentialgleichung} + Gegeben: + \[ + J + = + \dot\gamma(0) = \begin{pmatrix} 0 & -1 \\ 1 & \phantom-0 \end{pmatrix} + \] + Gesucht: + \[ \dot \gamma (t) = J \gamma(t) \qquad \gamma \in \gSO2 \] + \[ \Rightarrow \gamma(t) = \exp(Jt) \gamma(0) = \exp(Jt) \] + \end{block} + } + \end{column} + \begin{column}{0.48\textwidth} + \uncover<2->{ + \begin{block}{Lie-Algebra} + Potenzen von $J$: + \begin{align*} + J^2 &= -I & + J^3 &= -J & + J^4 &= I & + \ldots + \end{align*} + \end{block} + } + \end{column} + \end{columns} +\uncover<3->{ + Folglich: + \begin{align*} + \exp(Jt) + &= I + Jt + + J^2\frac{t^2}{2!} + + J^3\frac{t^3}{3!} + + J^4\frac{t^4}{4!} + + J^5\frac{t^5}{5!} + + \ldots \\ + &= \begin{pmatrix} + \vspace*{3pt} + 1 - \frac{t^2}{2} + \frac{t^4}{4!} - \ldots + & + -t + \frac{t^3}{3!} - \frac{t^5}{5!} + \ldots + \\ + t - \frac{t^3}{3!} + \frac{t^5}{5!} - \ldots + & + 1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \ldots + \end{pmatrix} + = \begin{pmatrix} \cos t & -\sin t \\ \sin t & \phantom-\cos t \end{pmatrix} - \right|_{ t = 0} - \\ - = - & - \begin{pmatrix} -\sin0 & -\cos0 \\ \phantom-\cos0 & -\sin0 \end{pmatrix} - = - \begin{pmatrix} 0 & -1 \\ 1 & \phantom-0 \end{pmatrix} - \end{align*} - \end{block} -\end{column} -\begin{column}{0.48\textwidth} - \begin{block}{Lie-Algebra} - Darstellung von \aso2: - \begin{align*} - \mathbb R - &\to - \aso2 - \\ - t - &\mapsto - \begin{pmatrix} - 0 & -t \\ - t & \phantom-0 - \end{pmatrix} - \end{align*} - \end{block} -\end{column} -\end{columns} -\end{frame} - - -\begin{frame}[t] -\setlength{\abovedisplayskip}{5pt} -\setlength{\belowdisplayskip}{5pt} -\frametitle{Von der Lie-Algebra zur -Gruppe} -\vspace{-20pt} -\begin{columns}[t,onlytextwidth] -\begin{column}{0.48\textwidth} - \begin{block}{Differentialgleichung} - Gegeben: - \[ - A - = - \dot\gamma(0) = \begin{pmatrix} 0 & -1 \\ 1 & \phantom-0 \end{pmatrix} - \] - Gesucht: - \[ \dot \gamma (t) = \gamma(t) A \qquad \gamma \in \gSO2 \] - \[ \Rightarrow \gamma(t) = \exp(At) \gamma(0) = \exp(At) \] - \end{block} -\end{column} -\begin{column}{0.48\textwidth} - \begin{block}{Lie-Algebra} - Potenzen von A: - \begin{align*} - A^2 &= -I & - A^3 &= -A & - A^4 &= I & - \ldots - \end{align*} - \end{block} -\end{column} -\end{columns} -Folglich: -\begin{align*} - \exp(At) - &= I + At - + A^2\frac{t^2}{2!} - + A^3\frac{t^3}{3!} - + A^4\frac{t^4}{4!} - + A^5\frac{t^5}{5!} - + \ldots \\ - &= \begin{pmatrix} - \vspace*{3pt} - 1 - \frac{t^2}{2} + \frac{t^4}{4!} - \ldots - & - -t + \frac{t^3}{3!} - \frac{t^5}{5!} + \ldots - \\ - t - \frac{t^3}{3!} + \frac{t^5}{5!} - \ldots - & - 1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \ldots - \end{pmatrix} - = - \begin{pmatrix} - \cos t & -\sin t \\ - \sin t & \phantom-\cos t - \end{pmatrix} -\end{align*} - + \end{align*} + } \end{frame} \egroup diff --git a/vorlesungen/slides/10/taylor.tex b/vorlesungen/slides/10/taylor.tex index 920470f..8c71965 100644 --- a/vorlesungen/slides/10/taylor.tex +++ b/vorlesungen/slides/10/taylor.tex @@ -10,12 +10,19 @@ \begin{frame}[t] \setlength{\abovedisplayskip}{5pt} \setlength{\belowdisplayskip}{5pt} - \frametitle{Beispiel $\sin x$} - \vspace{-20pt} - %\onslide<+-> - \begin{block}{Taylor-Approximationen von $\sin x$} + \frametitle{Beispiel $\sin(x)$} + \ifthenelse{\boolean{presentation}}{\vspace{-20pt}}{\vspace{-8pt}} + \begin{block}{Taylor-Approximationen von $\sin(x)$} \begin{align*} - p_n(x) + p_{ + \only<1>{0} + \only<2>{1} + \only<3>{2} + \only<4>{3} + \only<5>{4} + \only<6>{5} + \only<7->{n} + }(x) &= \uncover<1->{0} \uncover<2->{+ x} @@ -37,15 +44,15 @@ \draw[domain=-4:4, samples=50, smooth, blue] plot ({\x}, {sin(180/3.1415968*\x)}) node[above right] {$\sin(x)$}; - \uncover<1>{ + \uncover<1|handout:0>{ \draw[domain=-4:4, samples=2, smooth, red] plot ({\x}, {0}) node[above right] {$p_0(x)$};} - \uncover<2>{ + \uncover<2|handout:0>{ \draw[domain=-1.5:1.5, samples=2, smooth, red] plot ({\x}, {\x}) node[below right] {$p_1(x)$};} - \uncover<3>{ + \uncover<3|handout:0>{ \draw[domain=-1.5:1.5, samples=2, smooth, red] plot ({\x}, {\x}) node[below right] {$p_2(x)$};} @@ -53,19 +60,19 @@ \draw[domain=-3:3, samples=50, smooth, red] plot ({\x}, {\x - \x*\x*\x/6}) node[above right] {$p_3(x)$};} - \uncover<5>{ + \uncover<5|handout:0>{ \draw[domain=-3:3, samples=50, smooth, red] plot ({\x}, {\x - \x*\x*\x/6}) node[above right] {$p_4(x)$};} - \uncover<6>{ + \uncover<6|handout:0>{ \draw[domain=-3.9:3.9, samples=50, smooth, red] plot ({\x}, {\x - \x*\x*\x/6 + \x*\x*\x*\x*\x/120}) node[below right] {$p_5(x)$};} - \uncover<7>{ + \uncover<7|handout:0>{ \draw[domain=-3.9:3.9, samples=50, smooth, red] plot ({\x}, {\x - \x*\x*\x/6 + \x*\x*\x*\x*\x/120}) node[below right] {$p_6(x)$};} - \uncover<8->{ + \uncover<8-|handout:0>{ \draw[domain=-4:4, samples=50, smooth, red] plot ({\x}, {\x - \x*\x*\x/6 + \x*\x*\x*\x*\x/120 - \x*\x*\x*\x*\x*\x*\x/5040}) @@ -74,121 +81,134 @@ \end{center} \end{frame} - \begin{frame}[t] -\setlength{\abovedisplayskip}{5pt} -\setlength{\belowdisplayskip}{5pt} -\frametitle{Taylor-Reihen} -\vspace{-20pt} -\onslide<+-> - \begin{block}{Polynom-Approximationen von $f(t)$} - \vspace{-15pt} - \begin{align*} - p_n(t) - &= - f(0) - + f'(0) t - + f''(0)\frac{t^2}{2} - + f^{(3)}(0)\frac{t^3}{3!} - + \ldots - + f^{(n)}(0) \frac{t^n}{n!} - = - \sum_{k=0}^{n} f^{(k)} \frac{t^k}{k!} - \end{align*} - \end{block} - \begin{block}{Die ersten $n$ Ableitungen von $f(0)$ und $p_n(0)$ sind gleich!} - \vspace{-15pt} + \setlength{\abovedisplayskip}{5pt} + \setlength{\belowdisplayskip}{5pt} + \frametitle{Taylor-Reihen} + \ifthenelse{\boolean{presentation}}{\vspace{-20pt}}{\vspace{-8pt}} + \begin{block}{Polynom-Approximationen von $f(t)$} + \begin{align*} + p_n(t) + &= + f(0) + \uncover<2->{ + f' (0) t } + \uncover<3->{ + f''(0)\frac{t^2}{2} } + \uncover<4->{ + \ldots + f^{(n)}(0) \frac{t^n}{n!} } + \uncover<5->{ = \sum_{k=0}^{n} f^{(k)} \frac{t^k}{k!} } + \end{align*} + \end{block} + \uncover<6->{ + \begin{block}{Erste $n$ Ableitungen von $f(0)$ und $p_n(0)$ sind gleich!}} \begin{align*} - p'_n(t) - &= - f'(0) - + f''(0)t - + f^{(3)}(0) \frac{t^2}{2!} - + \mathcal O(t^3) - &\Rightarrow&& - p'_n(0) = f'(0) + \uncover<6->{ p'_n(t) } + & + \uncover<7->{ + = f'(0) + + f''(0)t + + \mathcal O(t^2) + } + &\uncover<8->{\Rightarrow}&& + \uncover<8->{p'_n(0) = f'(0)} \\ - p''_n(0) - &= - f''(0) + f^{(3)}(0)t + \ldots + f^{(n)}(0) \frac{t^{n-2}}{(n-2)!} - &\Rightarrow&& - p''_n(0) = f''(0) - \end{align*} - \end{block} - \begin{block}{Für unendlich lange Polynome stimmen alle Ableitungen überein!} - \vspace{-15pt} - \begin{align*} - \lim_{n\to \infty} p_n(t) - = - f(t) + \uncover<9->{ p''_n(t) } + & + \uncover<10->{ + = f''(0) + + \mathcal O(t) + } + &\uncover<11->{\Rightarrow}&& + \uncover<11->{ p''_n(0) = f''(0) } \end{align*} \end{block} + \uncover<12->{ + \begin{block}{Für alle praktisch relevanten Funktionen $f(t)$ gilt:} + \begin{align*} + \lim_{n\to \infty} p_n(t) + = + f(t) + \end{align*} + \end{block} + } \end{frame} \begin{frame}[t] \setlength{\abovedisplayskip}{5pt} \setlength{\belowdisplayskip}{5pt} - \frametitle{Beispiel $\exp x$} - \vspace{-20pt} - %\onslide<+-> - \begin{block}{Taylor-Approximationen von $\exp x$} + \frametitle{Beispiel $e^t$} + \ifthenelse{\boolean{presentation}}{\vspace{-20pt}}{\vspace{-8pt}} + \begin{block}{Taylor-Approximationen von $e^{at}$} \begin{align*} - p_n(x) - = + p_{ + \only<1>{0} + \only<2>{1} + \only<3>{2} + \only<4>{3} + \only<5>{4} + \only<6>{5} + \only<7->{n} + }(t) + &= 1 - \uncover<1->{+ x} - \uncover<2->{+ \frac{x^2}{2}} - \uncover<3->{+ \frac{x^3}{3!}} - \uncover<4->{+ \frac{x^4}{4!}} - \uncover<5->{+ \frac{x^5}{5!}} - \uncover<6->{+ \frac{x^6}{6!}} - \uncover<7->{+ \ldots - = \sum_{k=0}^{n} \frac{x^k}{k!}} + \uncover<2->{+ a t} + \uncover<3->{+ a^2 \frac{t^2}{2}} + \uncover<4->{+ a^3 \frac{t^3}{3!}} + \uncover<5->{+ a^4 \frac{t^4}{4!}} + \uncover<6->{+ a^5 \frac{t^5}{5!}} + \uncover<7->{+ a^6 \frac{t^6}{6!}} + \uncover<8->{+ \ldots + = \sum_{k=0}^{n} a^k \frac{t^k}{k!}} + \\ + & + \uncover<9->{= \exp(at)} \end{align*} \end{block} \begin{center} \begin{tikzpicture}[>=latex,thick,scale=1.3] - \draw[->] (-4.0, 0.0) -- (4.0,0.0) coordinate[label=$x$]; + \draw[->] (-4.0, 0.0) -- (4.0,0.0) coordinate[label=$t$]; \draw[->] ( 0.0,-0.5) -- (0.0,2.5); \clip (-3,-0.5) rectangle (3,2.5); \draw[domain=-4:1, samples=50, smooth, blue] plot ({\x}, {exp(\x)}) - node[above right] {$\exp(x)$}; - \uncover<1>{ + node[above right] {$\exp(t)$}; + \uncover<1|handout:0>{ + \draw[domain=-4:4, samples=12, smooth, red] + plot ({\x}, {1}) + node[below right] {$p_0(t)$};} + \uncover<2|handout:0>{ \draw[domain=-4:1.5, samples=10, smooth, red] - plot ({\x}, {1 + \x}) - node[below right] {$p_1(x)$};} - \uncover<2>{ + plot ({\x}, {1 + \x}) + node[below right] {$p_1(t)$};} + \uncover<3|handout:0>{ \draw[domain=-4:1, samples=50, smooth, red] plot ({\x}, {1 + \x + \x*\x/2}) - node[below right] {$p_2(x)$};} - \uncover<3>{ + node[below right] {$p_2(t)$};} + \uncover<4>{ \draw[domain=-4:1, samples=50, smooth, red] plot ({\x}, {1 + \x + \x*\x/2 + \x*\x*\x/6}) - node[below right] {$p_3(x)$};} - \uncover<4>{ + node[below right] {$p_3(t)$};} + \uncover<5|handout:0>{ \draw[domain=-4:0.9, samples=50, smooth, red] plot ({\x}, {1 + \x + \x*\x/2 + \x*\x*\x/6 + \x*\x*\x*\x/24}) - node[below left] {$p_4(x)$};} - \uncover<5>{ + node[below left] {$p_4(t)$};} + \uncover<6|handout:0>{ \draw[domain=-4:0.9, samples=50, smooth, red] plot ({\x}, {1 + \x + \x*\x/2 + \x*\x*\x/6 + \x*\x*\x*\x/24 + \x*\x*\x*\x*\x/120}) - node[below left] {$p_5(x)$};} - \uncover<6>{ + node[below left] {$p_5(t)$};} + \uncover<7|handout:0>{ \draw[domain=-4:0.9, samples=50, smooth, red] plot ({\x}, {1 + \x + \x*\x/2 + \x*\x*\x/6 + \x*\x*\x*\x/24 + \x*\x*\x*\x*\x/120 + \x*\x*\x*\x*\x*\x/720}) - node[below left] {$p_6(x)$};} - \uncover<7>{ + node[below left] {$p_6(t)$};} + \uncover<8-|handout:0>{ \draw[domain=-4:0.9, samples=50, smooth, red] plot ({\x}, {1 + \x + \x*\x/2 + \x*\x*\x/6 + \x*\x*\x*\x/24 + \x*\x*\x*\x*\x/120 + \x*\x*\x*\x*\x*\x/720 + \x*\x*\x*\x*\x*\x*\x/5040}) - node[below left] {$p_7(x)$};} + node[below left] {$p_7(t)$};} \end{tikzpicture} \end{center} \end{frame} diff --git a/vorlesungen/slides/10/vektorfelder.mp b/vorlesungen/slides/10/vektorfelder.mp index f488327..e63b2d5 100644 --- a/vorlesungen/slides/10/vektorfelder.mp +++ b/vorlesungen/slides/10/vektorfelder.mp @@ -48,17 +48,17 @@ drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10)); label.top(btex $x_1$ etex, z2 shifted (10,0)); label.rt(btex $x_2$ etex, z4 shifted (0,10)); -% Draw circles -for x = 0.2 step 0.2 until 1.4: - path p; - p = (x,0); - for a = 5 step 5 until 355: - p := p--(x*cosd(a), x*sind(a)); - endfor; - p := p--cycle; - pickup pencircle scaled 1pt; - draw p scaled unit withcolor red; -endfor; +% % Draw circles +% for x = 0.2 step 0.2 until 1.4: +% path p; +% p = (x,0); +% for a = 5 step 5 until 355: +% p := p--(x*cosd(a), x*sind(a)); +% endfor; +% p := p--cycle; +% pickup pencircle scaled 1pt; +% draw p scaled unit withcolor red; +% endfor; % Define DGL def dglField(expr x, y) = @@ -66,6 +66,10 @@ def dglField(expr x, y) = (-y, x) enddef; +pair A; +A := (1, 0); +draw A scaled unit withpen pencircle scaled 8bp withcolor red; + % Draw arrows for each grid point pickup pencircle scaled 0.5pt; for x = -1.5 step 0.1 until 1.55: @@ -78,11 +82,9 @@ endfor; endfig; - - % % Vektorfeld in der Ebene mit Lösungskurve -% X \in sl(2, R) +% Euler(1) % beginfig(2) @@ -101,18 +103,28 @@ drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10)); label.top(btex $x_1$ etex, z2 shifted (10,0)); label.rt(btex $x_2$ etex, z4 shifted (0,10)); -% Draw flow lines -for y = -1.4 step 0.2 until 1.4: +def dglField(expr x, y) = + (-y, x) +enddef; + +def dglFieldp(expr z) = + dglField(xpart z, ypart z) +enddef; + +def curve(expr z, l, s) = path p; - p = (-1.5,y) -- (1.5, y); - pickup pencircle scaled 1pt; + p := z; + for t = 0 step 1 until l: + p := p--((point (length p) of p) shifted (s * dglFieldp(point (length p) of p))); + endfor; draw p scaled unit withcolor red; -endfor; - -def dglField(expr x, y) = - (y, 0) enddef; +pair A; +A := (1, 0); +draw A scaled unit withpen pencircle scaled 8bp withcolor red; +curve(A, 0, 1); + % Draw arrows for each grid point pickup pencircle scaled 0.5pt; for x = -1.5 step 0.1 until 1.55: @@ -125,12 +137,9 @@ endfor; endfig; - - - % % Vektorfeld in der Ebene mit Lösungskurve -% Y \in sl(2, R) +% Euler(2) % beginfig(3) @@ -149,42 +158,82 @@ drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10)); label.top(btex $x_1$ etex, z2 shifted (10,0)); label.rt(btex $x_2$ etex, z4 shifted (0,10)); -% Draw flow lines -for x = -1.4 step 0.2 until 1.4: +def dglField(expr x, y) = + (-y, x) +enddef; + +def dglFieldp(expr z) = + dglField(xpart z, ypart z) +enddef; + +def curve(expr z, l, s) = path p; - p = (x, -1.5) -- (x, 1.5); - pickup pencircle scaled 1pt; + p := z; + for t = 0 step 1 until l: + p := p--((point (length p) of p) shifted (s * dglFieldp(point (length p) of p))); + endfor; draw p scaled unit withcolor red; +enddef; + +pair A; +A := (1, 0); +draw A scaled unit withpen pencircle scaled 8bp withcolor red; +curve(A, 1, 0.5); + +% Draw arrows for each grid point +pickup pencircle scaled 0.5pt; +for x = -1.5 step 0.1 until 1.55: + for y = -1.5 step 0.1 until 1.55: + drawarrow ((x, y) * unit) + --(((x,y) * unit) shifted (8 * dglField(x,y))) + withcolor blue; + endfor; endfor; -def dglField(expr x, y) = - (0, x) -enddef; +endfig; -% def dglFieldp(expr z) = -% dglField(xpart z, ypart z) -% enddef; -% -% def curve(expr z, l) = -% path p; -% p := z; -% for t = 0 step 1 until l: -% p := p--((point (length p) of p) shifted (0.01 * dglFieldp(point (length p) of p))); -% endfor; -% draw p scaled unit withcolor red; -% enddef; % -% numeric outerlength; -% outerlength = 200; -% curve(( 0.1, 0), outerlength); -% curve(( 0.2, 0), outerlength); +% Vektorfeld in der Ebene mit Lösungskurve +% Euler(3) % -% numeric innerlength; -% innerlength = 500; -% -% for a = 0 step 30 until 330: -% curve(0.05 * (cosd(a), sind(a)), innerlength); -% endfor; +beginfig(4) + +numeric unit; +unit := 150; + +z0 = ( 0, 0); +z1 = (-1.5, 0) * unit; +z2 = ( 1.5, 0) * unit; +z3 = ( 0, -1.5) * unit; +z4 = ( 0, 1.5) * unit; + +pickup pencircle scaled 1pt; +drawarrow (z1 shifted (-10,0))--(z2 shifted (10,0)); +drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10)); +label.top(btex $x_1$ etex, z2 shifted (10,0)); +label.rt(btex $x_2$ etex, z4 shifted (0,10)); + +def dglField(expr x, y) = + (-y, x) +enddef; + +def dglFieldp(expr z) = + dglField(xpart z, ypart z) +enddef; + +def curve(expr z, l, s) = + path p; + p := z; + for t = 0 step 1 until l: + p := p--((point (length p) of p) shifted (s * dglFieldp(point (length p) of p))); + endfor; + draw p scaled unit withcolor red; +enddef; + +pair A; +A := (1, 0); +draw A scaled unit withpen pencircle scaled 8bp withcolor red; +curve(A, 3, 0.25); % Draw arrows for each grid point pickup pencircle scaled 0.5pt; @@ -198,12 +247,11 @@ endfor; endfig; - % % Vektorfeld in der Ebene mit Lösungskurve -% H \in sl(2, R) +% Euler(4) % -beginfig(4) +beginfig(5) numeric unit; unit := 150; @@ -215,40 +263,88 @@ z3 = ( 0, -1.5) * unit; z4 = ( 0, 1.5) * unit; pickup pencircle scaled 1pt; -drawarrow (z1 shifted (-25,0))--(z2 shifted (25,0)); +drawarrow (z1 shifted (-10,0))--(z2 shifted (10,0)); drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10)); -label.top(btex $x_1$ etex, z2 shifted (25,0)); +label.top(btex $x_1$ etex, z2 shifted (10,0)); label.rt(btex $x_2$ etex, z4 shifted (0,10)); def dglField(expr x, y) = - (x, -y) + (-y, x) enddef; def dglFieldp(expr z) = dglField(xpart z, ypart z) enddef; -def curve(expr z, l) = +def curve(expr z, l, s) = path p; p := z; for t = 0 step 1 until l: - p := p--((point (length p) of p) shifted (0.01 * dglFieldp(point (length p) of p))); + p := p--((point (length p) of p) shifted (s * dglFieldp(point (length p) of p))); endfor; draw p scaled unit withcolor red; enddef; -for i = -1 step 2 until 1: - for k = -1 step 2 until 1: - curve((1.3 * i, 1.5 * k), 18); - curve((1.1 * i, 1.5 * k), 35); - curve((0.9 * i, 1.5 * k), 55); - curve((0.7 * i, 1.5 * k), 80); - curve((0.5 * i, 1.5 * k), 114); - curve((0.3 * i, 1.5 * k), 165); - curve((0.1 * i, 1.5 * k), 275); +pair A; +A := (1, 0); +draw A scaled unit withpen pencircle scaled 8bp withcolor red; +curve(A, 7, 0.125); + +% Draw arrows for each grid point +pickup pencircle scaled 0.5pt; +for x = -1.5 step 0.1 until 1.55: + for y = -1.5 step 0.1 until 1.55: + drawarrow ((x, y) * unit) + --(((x,y) * unit) shifted (8 * dglField(x,y))) + withcolor blue; endfor; endfor; +endfig; + +% +% Vektorfeld in der Ebene mit Lösungskurve +% Euler(5) +% +beginfig(6) + +numeric unit; +unit := 150; + +z0 = ( 0, 0); +z1 = (-1.5, 0) * unit; +z2 = ( 1.5, 0) * unit; +z3 = ( 0, -1.5) * unit; +z4 = ( 0, 1.5) * unit; + +pickup pencircle scaled 1pt; +drawarrow (z1 shifted (-10,0))--(z2 shifted (10,0)); +drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10)); +label.top(btex $x_1$ etex, z2 shifted (10,0)); +label.rt(btex $x_2$ etex, z4 shifted (0,10)); + +def dglField(expr x, y) = + (-y, x) +enddef; + +def dglFieldp(expr z) = + dglField(xpart z, ypart z) +enddef; + +def curve(expr z, l, s) = + path p; + p := z; + for t = 0 step 1 until l: + p := p--((point (length p) of p) shifted (s * dglFieldp(point (length p) of p))); + endfor; + draw p scaled unit withcolor red; +enddef; + +pair A; +A := (1, 0); +draw A scaled unit withpen pencircle scaled 8bp withcolor red; +curve(A, 99, 0.01); + % Draw arrows for each grid point pickup pencircle scaled 0.5pt; for x = -1.5 step 0.1 until 1.55: @@ -262,5 +358,4 @@ endfor; endfig; - end; diff --git a/vorlesungen/slides/10/vektorfelder.tex b/vorlesungen/slides/10/vektorfelder.tex new file mode 100644 index 0000000..3ba7cda --- /dev/null +++ b/vorlesungen/slides/10/vektorfelder.tex @@ -0,0 +1,82 @@ +% +% iterativ.tex -- Iterative Approximation in \dot x = J x +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% Erstellt durch Roy Seitz +% +% !TeX spellcheck = de_CH +\bgroup +\begin{frame}[t] + \setlength{\abovedisplayskip}{5pt} + \setlength{\belowdisplayskip}{5pt} + \frametitle{Als Strömungsfeld} + \vspace{-20pt} + \begin{columns}[t,onlytextwidth] + \begin{column}{0.48\textwidth} + \vfil + \only<1|handout:0>{ + \includegraphics[width=\linewidth,keepaspectratio] + {../slides/10/vektorfelder-1.pdf} + } + \only<2|handout:0>{ + \includegraphics[width=\linewidth,keepaspectratio] + {../slides/10/vektorfelder-2.pdf} + } + \only<3>{ + \includegraphics[width=\linewidth,keepaspectratio] + {../slides/10/vektorfelder-3.pdf} + } + \only<4|handout:0>{ + \includegraphics[width=\linewidth,keepaspectratio] + {../slides/10/vektorfelder-4.pdf} + } + \only<5|handout:0>{ + \includegraphics[width=\linewidth,keepaspectratio] + {../slides/10/vektorfelder-5.pdf} + } + \only<6-|handout:0>{ + \includegraphics[width=\linewidth,keepaspectratio] + {../slides/10/vektorfelder-6.pdf} + } + \vfil + \end{column} + \begin{column}{0.48\textwidth} + \begin{block}{Differentialgleichung} + \[ + \dot x(t) = J x(t) + \quad + J = \begin{pmatrix} 0 & -1 \\ 1 & \phantom-0 \end{pmatrix} + \quad + x_0 = \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \] + \end{block} + + \only<2|handout:0>{ + Nach einem Schritt der Länge $t$: + \[ + x(t) = x_0 + \dot x t = x_0 + Jx_0t = (1 + Jt)x_0 + \] + } + + \only<3|handout:0>{ + Nach zwei Schritten der Länge $t/2$: + \[ + x(t) = \left(1 + \frac{Jt}{2}\right)^2x_0 + \] + } + + \only<4->{ + Nach n Schritten der Länge $t/n$: + \[ + x(t) = \left(1 + \frac{Jt}{n}\right)^nx_0 + \] + } + \only<6->{ + \[ + \lim_{n\to\infty}\left(1 + \frac{At}{n}\right)^n = \exp(At) + \] + } + \end{column} + \end{columns} +\end{frame} +\egroup |