From 8358d9bc031913305a52c6c2ab05184b89f7678f Mon Sep 17 00:00:00 2001 From: Roy Seitz Date: Sat, 17 Apr 2021 22:00:36 +0200 Subject: Slides erweitert. --- vorlesungen/slides/10/ableitung-exp.tex | 60 +++++++++ vorlesungen/slides/10/matrix-dgl.tex | 83 ++++++++++++ vorlesungen/slides/10/matrix-vektor-dgl.tex | 83 ------------ vorlesungen/slides/10/potenzreihenmethode.tex | 91 +++++++++++++ vorlesungen/slides/10/repetition.tex | 151 ++++++++++++++++++++++ vorlesungen/slides/10/so2.tex | 8 -- vorlesungen/slides/10/taylor.tex | 176 ++++++++++++++------------ 7 files changed, 483 insertions(+), 169 deletions(-) create mode 100644 vorlesungen/slides/10/ableitung-exp.tex create mode 100644 vorlesungen/slides/10/matrix-dgl.tex delete mode 100644 vorlesungen/slides/10/matrix-vektor-dgl.tex create mode 100644 vorlesungen/slides/10/potenzreihenmethode.tex create mode 100644 vorlesungen/slides/10/repetition.tex (limited to 'vorlesungen/slides/10') 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/matrix-dgl.tex b/vorlesungen/slides/10/matrix-dgl.tex new file mode 100644 index 0000000..ae68fb1 --- /dev/null +++ b/vorlesungen/slides/10/matrix-dgl.tex @@ -0,0 +1,83 @@ +% +% matrix-dgl.tex -- Matrix-Differentialgleichungen +% +% (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{1.~Ordnung mit Skalaren} + \vspace{-20pt} + \begin{columns}[t,onlytextwidth] + \begin{column}{0.48\textwidth} + \begin{block}{Aufgabe} + Sei $a, x(t), x_0 \in \mathbb R$, + \[ + \dot x(t) = ax(t), + \quad + x(0) = x_0 + \] + \end{block} + \begin{block}{Potenzreihen-Ansatz} + Sei $a_k \in \mathbb R$, + \[ + x(t) = a_0 + a_1t + a_2t^2 + a_3t^3 \ldots + \] + \end{block} + \end{column} + \begin{column}{0.48\textwidth} + \begin{block}{Lösung} + Einsetzen in DGL, Koeffizientenvergleich liefert + \[ x(t) = \exp(at) \, x_0, \] + wobei + \begin{align*} + \exp(at) + &= 1 + at + \frac{a^2t^2}{2} + \frac{a^3t^3}{3!} + \ldots \\ + &{\color{gray}(= e^{at}.)} + \end{align*} + \end{block} + \end{column} + \end{columns} +\end{frame} + +\begin{frame}[t] + \setlength{\abovedisplayskip}{5pt} + \setlength{\belowdisplayskip}{5pt} + \frametitle{1.~Ordnung mit Matrizen} + \vspace{-20pt} + \begin{columns}[t,onlytextwidth] + \begin{column}{0.48\textwidth} + \begin{block}{Aufgabe} + Sei $A \in M_n$, $x(t), x_0 \in \mathbb R^n$, + \[ + \dot x(t) = Ax(t), + \quad + x(0) = x_0 + \] + \end{block} + \begin{block}{Potenzreihen-Ansatz} + Sei $A_k \in \mathbb M_n$, + \[ + x(t) = A_0 + A_1t + A_2t^2 + A_3t^3 \ldots + \] + \end{block} + \end{column} + \begin{column}{0.48\textwidth} + \begin{block}{Lösung} + Einsetzen in DGL, Koeffizientenvergleich liefert + \[ x(t) = \exp(At) \, x_0, \] + wobei + \[ + \exp(At) + = 1 + At + \frac{A^2t^2}{2} + \frac{A^3t^3}{3!} + \ldots + \] + \end{block} + \end{column} + \end{columns} +\end{frame} + +\egroup diff --git a/vorlesungen/slides/10/matrix-vektor-dgl.tex b/vorlesungen/slides/10/matrix-vektor-dgl.tex deleted file mode 100644 index f7bd995..0000000 --- a/vorlesungen/slides/10/matrix-vektor-dgl.tex +++ /dev/null @@ -1,83 +0,0 @@ -% -% matrix-vektor-dgl.tex -- DGL mit Matrix-Koeffizienten und Vektor-Variablen -% -% (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{1.~Ordnung mit Skalaren} - \vspace{-20pt} - \begin{columns}[t,onlytextwidth] - \begin{column}{0.48\textwidth} - \begin{block}{Aufgabe} - Sei $a, x(t), x_0 \in \mathbb R$, - \[ - \dot x(t) = ax(t), - \quad - x(0) = x_0 - \] - \end{block} - \begin{block}{Potenzreihen-Ansatz} - Sei $a_k \in \mathbb R$, - \[ - x(t) = a_0 + a_1t + a_2t^2 + a_3t^3 \ldots - \] - \end{block} - \end{column} - \begin{column}{0.48\textwidth} - \begin{block}{Lösung} - Einsetzen in DGL, Koeffizientenvergleich liefert - \[ x(t) = \exp(at) \, x_0, \] - wobei - \begin{align*} - \exp(at) - &= 1 + at + \frac{a^2t^2}{2} + \frac{a^3t^3}{3!} + \ldots \\ - &{\color{gray}(= e^{at}.)} - \end{align*} - \end{block} - \end{column} - \end{columns} -\end{frame} - -\begin{frame}[t] - \setlength{\abovedisplayskip}{5pt} - \setlength{\belowdisplayskip}{5pt} - \frametitle{1.~Ordnung mit Matrizen} - \vspace{-20pt} - \begin{columns}[t,onlytextwidth] - \begin{column}{0.48\textwidth} - \begin{block}{Aufgabe} - Sei $A \in M_n$, $x(t), x_0 \in \mathbb R^n$, - \[ - \dot x(t) = Ax(t), - \quad - x(0) = x_0 - \] - \end{block} - \begin{block}{Potenzreihen-Ansatz} - Sei $A_k \in \mathbb M_n$, - \[ - x(t) = A_0 + A_1t + A_2t^2 + A_3t^3 \ldots - \] - \end{block} - \end{column} - \begin{column}{0.48\textwidth} - \begin{block}{Lösung} - Einsetzen in DGL, Koeffizientenvergleich liefert - \[ x(t) = \exp(At) \, x_0, \] - wobei - \[ - \exp(At) - = 1 + At + \frac{A^2t^2}{2} + \frac{A^3t^3}{3!} + \ldots - \] - \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..c45d47b --- /dev/null +++ b/vorlesungen/slides/10/repetition.tex @@ -0,0 +1,151 @@ +% +% 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{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 & t \\ 0 & 0 \end{pmatrix} } + \\ + \uncover<5->{ Y(t) } + & + \uncover<5->{= \begin{pmatrix} 0 & 0 \\ t & 0 \end{pmatrix} } + \\ + \uncover<6->{ H(t) } + & + \uncover<6->{= \begin{pmatrix} t & 0 \\ 0 & -t \end{pmatrix} } + \end{align*} + + \end{column} + \end{columns} +\end{frame} + +\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 Was bedeutet $\exp(At)$? + \end{itemize} + \end{block} +\end{frame} + +\egroup diff --git a/vorlesungen/slides/10/so2.tex b/vorlesungen/slides/10/so2.tex index e3f74ae..b63a67e 100644 --- a/vorlesungen/slides/10/so2.tex +++ b/vorlesungen/slides/10/so2.tex @@ -7,14 +7,6 @@ % !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)}} - -\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} diff --git a/vorlesungen/slides/10/taylor.tex b/vorlesungen/slides/10/taylor.tex index 920470f..25745f5 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 $\sin x$} + \frametitle{Beispiel $\sin(x)$} \vspace{-20pt} - %\onslide<+-> - \begin{block}{Taylor-Approximationen von $\sin x$} + \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} @@ -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} + \vspace{-20pt} + \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 $e^t$} +% \vspace{-20pt} + \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] {$\exp(x)$}; + node[above right] {$\exp(t)$}; \uncover<1>{ - \draw[domain=-4:1.5, samples=10, smooth, red] - plot ({\x}, {1 + \x}) - node[below right] {$p_1(x)$};} + \draw[domain=-4:4, samples=12, smooth, red] + plot ({\x}, {1}) + node[below right] {$p_0(t)$};} \uncover<2>{ + \draw[domain=-4:1.5, samples=10, smooth, red] + plot ({\x}, {1 + \x}) + node[below right] {$p_1(t)$};} + \uncover<3>{ \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>{ \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>{ \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>{ \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->{ \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} -- cgit v1.2.1