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hw6: done 1 2 4

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Claudio Maggioni (maggicl) 2020-05-27 17:10:54 +02:00
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@ -109,4 +109,65 @@ Q = \frac9{16} + \frac14 \cdot \frac14 = \frac{10}{16}\]
+ f\left(1\right)\right) = \frac7{16} - \frac12 \cdot \frac34 = \frac1{16} < \frac1{10}\]
Thus the solution using quadrature is $\frac{5}{8}$.
\section*{Question 4}
\[\begin{bmatrix} x_0^2 & x_0 & 1 \\x_1^2 & x_1 & 1 \\x_2^2 & x_2 & 1 \\x_3^2 &
x_3 & 1 \end{bmatrix} \begin{bmatrix}a \\b\\c\\\end{bmatrix} = \begin{bmatrix}
y_0 \\y_1\\y_2\\y_3\\\end{bmatrix}\]
\[\begin{bmatrix} x_0^2 & x_1^2 & x_2^2 & x_3^2 \\ x_0 & x_1 & x_2 & x_3 \\
1&1&1&1\end{bmatrix} \begin{bmatrix} x_0^2 & x_0 & 1 \\x_1^2 & x_1 & 1 \\x_2^2 & x_2 & 1 \\x_3^2 &
x_3 & 1 \end{bmatrix} \begin{bmatrix}a \\b\\c\\\end{bmatrix} = \begin{bmatrix} x_0^2 & x_1^2 & x_2^2 & x_3^2 \\ x_0 & x_1 & x_2 & x_3 \\
1&1&1&1\end{bmatrix} \begin{bmatrix} y_0 \\y_1\\y_2\\y_3\\\end{bmatrix}\]
\[\begin{bmatrix}18&8&6\\8&6&2\\6&2&4\\\end{bmatrix}\begin{bmatrix}a\\b\\c\\\end{bmatrix}=\begin{bmatrix}2\\0\\2\\\end{bmatrix}\]
We now use Gaussian \textit{ellimination} to solve the system:
\[
\begin{array}{@{}ccc|c@{}}
18&8&6&2\\
8&6&2&0\\
6&2&4&2\\
\end{array}
\qquad
\begin{array}{@{}ccc|c@{}}
1&\frac49 &\frac13 & \frac19\\
8& 6& 2 &0\\
6 &2 &4 &2\\
\end{array}
\qquad
\begin{array}{@{}ccc|c@{}}
1& \frac49& \frac13& \frac19\\
0& \frac{22}9& \frac{-2}3& \frac{-8}9\\
0& \frac{-2}3& 2& \frac43\\
\end{array}
\qquad
\begin{array}{@{}ccc|c@{}}
1& \frac49& \frac13& \frac19\\
0 &1 &\frac{-3}{11} & \frac{-4}{11}\\
0 & \frac{-2}{3} & 2 & \frac43\\
\end{array}
\]\[
\begin{array}{@{}ccc|c@{}}
1& 0& \frac5{11} &\frac3{11}\\
0 &1 &\frac{-3}{11}&\frac{-4}{11}\\
0 &0 &\frac{20}{11}&\frac{12}{11}\\
\end{array}
\qquad
\begin{array}{@{}ccc|c@{}}
1& 0 &\frac5{11} & \frac{3}{11}\\
0 &1 & \frac{-3}{11} & \frac{-4}{11}\\
0& 0& 1 &\frac35\\
\end{array}
\qquad
\begin{array}{@{}ccc|c@{}}
1 &0 &0 &0\\
0& 1 &0 &\frac{-1}5\\
0& 0& 1 &\frac35\\
\end{array}
\qquad
\begin{bmatrix}a\\b\\c\\\end{bmatrix}=\begin{bmatrix}0\\-\frac15\\\frac35\\\end{bmatrix}
\]
\end{document}