Mathematics, 29.02.2020 00:26 lindsaynielsen13
Fit a quadratic function of the form f(t)=c0+c1t+c2t2f(t)=c0+c1t+c2t2 to the data points (0,−5)(0,−5), (1,5)(1,5), (2,−3)(2,−3), (3,11)(3,11), using least squares.
Answers: 1
![A=\left[\begin{array}{ccc}1&0&0\\1&1&1\\1&2&4\\1&3&9\end{array}\right]](/tpl/images/0528/7717/6b69e.png)
, ![\vec B=\left[\begin{array}{c}-5\\5\\-3\\11\end{array}\right]](/tpl/images/0528/7717/55f78.png)
and ![\vec x=\left[\begin{array}{ccc}c_0\\c_1\\c_2\end{array}\right]](/tpl/images/0528/7717/5dbda.png)
![rref A=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\\0&0&0\end{array}\right]](/tpl/images/0528/7717/fe847.png)
![A^T=\left[\begin{array}{cccc}1&1&1&1\\0&1&2&3\\0&1&4&2\end{array}\right]](/tpl/images/0528/7717/e86d1.png)
![=\left( \left[\begin{array}{cccc}1&1&1&1\\0&1&2&3\\0&1&4&2\end{array}\right]\left[\begin{array}{ccc}1&0&0\\1&1&1\\1&2&4\\1&3&9\end{array}\right]\right)^{-1} \left(\left[\begin{array}{cccc}1&1&1&1\\0&1&2&3\\0&1&4&2\end{array}\right]\left[\begin{array}{c}-5\\5\\-3\\11\end{array}\right] \right)](/tpl/images/0528/7717/f16bd.png)
![=\left(\left[\begin{array}{ccc}4&6&14\\6&14&36\\7&15&35\end{array}\right] \right)^{-1}\left[\begin{array}{c}8\\32\\15\end{array}\right]](/tpl/images/0528/7717/1cde5.png)
![=\left[\begin{array}{ccc}\frac{5}{6}&0&-\frac{1}{3}\ \\-\frac{7}{10} &-\frac{7}{10}&1\\\frac{2}{15} &\frac{3}{10} &-\frac{1}{3}\end{array}\right] \left[\begin{array}{c}8\\32\\15\end{array}\right]](/tpl/images/0528/7717/5afca.png)
![=\left[\begin{array}{c}\frac{5}{3} \\-13\\\frac{17}{3} \end{array}\right]](/tpl/images/0528/7717/4b078.png)
in the given quadratic equation
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Fit a quadratic function of the form f(t)=c0+c1t+c2t2f(t)=c0+c1t+c2t2 to the data points (0,−5)(0,−5...
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