(Solved): Consider the data points in Example 6. We want to fit the polynomial \( y=c_{1} x+c_{2} x^{2}+c_{3 ...

Consider the data points in Example 6. We want to fit the polynomial \( y=c_{1} x+c_{2} x^{2}+c_{3} x^{3} \) to the data. The corresponding matrix \( X \) has the form: Hint: substitute the coordinates of the points into the equation and determine the corresponding matrix \( X \). \[ \begin{aligned} X &=\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 2 & 4 \end{array}\right] \\ X &=\left[\begin{array}{lll} 0 & 0 & 0 \\ 1 & 1 & 1 \\ 2 & 4 & 8 \\ 2 & 4 & 8 \end{array}\right] \\ X &=\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \\ 1 & 8 \\ 1 & 8 \end{array}\right] \end{aligned} \] Example 6. Fit \( y=c_{0}+c_{1} x+c_{2} x^{2} \) to the data \( \boldsymbol{\theta}=\left[\begin{array}{lll}1 & x_{1} & x_{1}^{2} \\ 1 & x_{2} & x_{2}^{2} \\ 1 & x_{3} & x_{3}^{2} \\ 1 & x_{4} & x_{4}^{2}\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 2 & 4\end{array}\right], \quad \mathbf{y}=\left[\begin{array}{l}1 \\ 0 \\ \frac{3}{2} \\ 2\end{array}\right] \) \( X^{T} X \mathbf{c}=X^{T} \mathbf{y} \Leftrightarrow\left[\begin{array}{ccc}4 & 5 & 9 \\ 5 & 9 & 17 \\ 9 & 17 & 33\end{array}\right] \mathbf{c}=\left[\begin{array}{c}\frac{9}{2} \\ 7 \\ 14\end{array}\right] \) \( \boldsymbol{0}\left[\begin{array}{l}c_{0} \\ c_{1} \\ c_{2}\end{array}\right]=\mathbf{c}=\left[\begin{array}{r}1 \\ -\frac{19}{8} \\ \frac{11}{8}\end{array}\right] \Rightarrow\left[\begin{array}{c}1 \\ -2.375 \\ 1.375\end{array}\right] \) \( \Rightarrow \quad \) the model is \( y=1-\frac{19}{8} x+\frac{11}{8} x^{2} \)