From 8dad61d27a6998b82d51e8f94d6fedb87f48302c Mon Sep 17 00:00:00 2001 From: Andreas Tsouchlos Date: Wed, 14 Jan 2026 01:30:12 +0100 Subject: [PATCH] Finish solution for exercise 2a --- src/2026-01-16/presentation.tex | 61 ++++++++++++++++++++------------- 1 file changed, 37 insertions(+), 24 deletions(-) diff --git a/src/2026-01-16/presentation.tex b/src/2026-01-16/presentation.tex index 61f9b6b..5e05529 100644 --- a/src/2026-01-16/presentation.tex +++ b/src/2026-01-16/presentation.tex @@ -249,7 +249,16 @@ X = h_1(U,V) = U \\ Y = h_2(U,V) = \frac{V}{U} \end{array} - \right. \\[4mm] + \hspace{20mm} + \left(\begin{array}{l} + 0 < x \le 1 \Rightarrow 0 < u \le 1 \\ + 0 < y \le 1 \Rightarrow 0 < v \le u \le 1 + \end{array} + \right) + \right. + \end{gather*} + \vspace*{5mm} + \pause\begin{gather} \mathcal{J} = \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\[2mm] \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} @@ -258,32 +267,37 @@ 1 & 0 \\ - \frac{v}{u^2} & \frac{1}{u} \end{pmatrix} - \end{gather*} - \vspace{3mm} + \end{gather} \begin{align*} - f_{U,V}(u,v) = \lvert \text{det}(\mathcal{J}) \rvert \cdot f_{X,Y} \big(h_1(u,v),h_2(u,v)\big) - = \left\{ \begin{array}{ll} - \frac{1}{u} \cdot \left(u + \frac{v}{u}\right) &,\hspace{2mm} 0 < v \le u \le 1 \\ - 0 &,\hspace{2mm} \text{sonst} - \end{array} \right. - = \left\{ \begin{array}{ll} - 1 + \frac{v}{u^2} &,\hspace{2mm} 0 < v \le u \le 1 \\ - 0 &,\hspace{2mm} \text{sonst} - \end{array} \right. + f_{U,V}(u,v) &= \lvert \text{det}(\mathcal{J}) \rvert + \cdot f_{X,Y} \big(h_1(u,v),h_2(u,v)\big) + = \frac{1}{u} \cdot \left(u + \frac{v}{u}\right) + = 1 + \frac{v}{u^2}, && \hspace*{-20mm} 0 < v \le u \le 1 \\[3mm] \end{align*} - \vspace{3mm} - \begin{gather*} - \text{Für } 0 < v \le 1: \hspace{5mm} f_V(v) - = \int_{-\infty}^{\infty} f_{U,V}(u,v) du - = \int_{v}^{1} 1 + \frac{v}{u^2} dx + \vspace*{-22mm} + \pause\begin{align*} + f_V(v) &= \int_{-\infty}^{\infty} f_{U,V}(u,v) du + = \int_{v}^{1} 1 + \frac{v}{u^2} du = \left[ u - \frac{v}{u} \right]_v^1 - = 2(1-v) \\[3mm] - f_Z(z) = \left\{\begin{array}{ll} - 2(1-z) \hspace{3mm}&,\hspace{3mm} 0 < z \le 1 \\ - 0 \hspace{3mm}&,\hspace{3mm} \text{sonst}\\ - \end{array}\right. + = 2(1-v), && \hspace*{-20mm} 0 < v \le 1 + \end{align*} + \vspace{5mm} + \pause\begin{gather*} + f_Z(z) = \left\{\begin{array}{ll} + 2(1-z) \hspace{3mm}&,\hspace{3mm} 0 < z \le 1 \\ + 0 \hspace{3mm}&,\hspace{3mm} \text{sonst}\\ + \end{array}\right. \end{gather*} - \pause \item Verwenden Sie einen alternativen Ansatz zur Berechnung der + \end{enumerate} + % tex-fmt: on +\end{frame} + +\begin{frame} + \frametitle{Aufgabe 2: Transformationssatz für 2D-Dichten} + + \begin{enumerate}[a{)}] + \setcounter{enumi}{1} + \item Verwenden Sie einen alternativen Ansatz zur Berechnung der Dichte. Hinweis: Beginnen Sie mit $P (Z \le z) = \ldots$ \pause\begin{align*} \end{align*} @@ -291,7 +305,6 @@ \pause\begin{align*} \end{align*} \end{enumerate} - % tex-fmt: on \end{frame} \end{document}