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93 \frametitle{Outline} |
93 \frametitle{Outline} |
94 \tableofcontents[currentsection,currentsubsection] |
94 \tableofcontents[currentsection,currentsubsection] |
95 \end{frame} |
95 \end{frame} |
96 } |
96 } |
97 |
97 |
98 %%\AtBeginSection[] |
98 \AtBeginSection[] |
99 %%{ |
99 { |
100 %%\begin{frame}<beamer> |
100 \begin{frame}<beamer> |
101 %% \frametitle{Outline} |
101 \frametitle{Outline} |
102 %% \tableofcontents[currentsection,currentsubsection] |
102 \tableofcontents[currentsection,currentsubsection] |
103 %%\end{frame} |
103 \end{frame} |
104 %%} |
104 } |
105 |
105 |
106 % If you wish to uncover everything in a step-wise fashion, uncomment |
106 % If you wish to uncover everything in a step-wise fashion, uncomment |
107 % the following command: |
107 % the following command: |
108 %\beamerdefaultoverlayspecification{<+->} |
108 %\beamerdefaultoverlayspecification{<+->} |
109 |
109 |
181 [-0.52532209, -0.08675134, -0.81649658], |
181 [-0.52532209, -0.08675134, -0.81649658], |
182 [-0.8186735 , 0.61232756, 0.40824829]])) |
182 [-0.8186735 , 0.61232756, 0.40824829]])) |
183 \end{lstlisting} |
183 \end{lstlisting} |
184 \end{frame} |
184 \end{frame} |
185 |
185 |
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186 \section{Solving linear equations} |
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187 \begin{frame}[fragile] |
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188 \frametitle{Solution of equations} |
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189 Example problem: Consider the set of equations |
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190 \begin{align*} |
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191 3x + 2y - z & = 1 \\ |
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192 2x - 2y + 4z & = -2 \\ |
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193 -x + \frac{1}{2}y -z & = 0 |
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194 \end{align*} |
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195 |
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196 To Solve this, |
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197 \begin{lstlisting} |
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198 In []: A = array([[3,2,-1],[2,-2,4],[-1, 0.5, -1]]) |
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199 In []: b = array([1, -2, 0]) |
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200 In []: x = linalg.solve(A, b) |
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201 In []: Ax = dot(A, x) |
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202 In []: allclose(Ax, b) |
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203 Out[]: True |
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204 \end{lstlisting} |
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205 \end{frame} |
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206 |
186 \end{document} |
207 \end{document} |