I'm trying to find least squares approximation $p(x)=c_1x+c_2x^2$ of $f(x)=xe^{x/2}$ in $[0,2]$.
Using the algorithm here, p.7.: http://www.math.niu.edu/~dattab/MATH435.2013/APPROXIMATION.pdf
I'm able to come up with a $3 \times 3$ matrix $S$ and $3 \times 1$ vector $b$. However this gives me as a solution three coefficients, even though I only have two. What do I need to do to get only two (i.e. no $c_0$)?
The approximation is $$ g(x) \approx c_{1} f_{1}(x) + c_{2} f_{2}(x) $$ over the domain $a < x < b$. Choose the method of least squares. Minimize the sums of the squares of residuals: $$ r^{2}(c) = \int_{a}^{b} \left(f(x) - c_{1} f_{1}(x) - c_{2} f_{2}(x) \right)^{2} dx. $$ This creates the linear system $$ \begin{align} % \mathbf{A} c &= G \\ % \left[ \begin{array}{cc} \int_{a}^{b} f_{1}(x) \times f_{1}(x) dx & \int_{a}^{b} f_{1}(x) \times f_{2}(x) dx \\ \int_{a}^{b} f_{2}(x) \times f_{1}(x) dx & \int_{a}^{b} f_{2}(x) \times f_{2}(x) dx \end{array} \right] % \left[ \begin{array}{c} c_{1} \\ c_{2} \end{array} \right] % &= % \left[ \begin{array}{c} \int_{a}^{b} f_{1}(x) \times g(x) dx \\ \int_{a}^{b} f_{2}(x) \times g(x) dx \end{array} \right] \end{align} $$ In this problem $g(x) = xe^{\frac{x}{2}}$, $f_{1}(x) = x$, and $f_{2}(x) = x^{2}.$
For the domain $a= 0$ and $b=2$, the linear system is $$ \left[ \begin{array}{cc} \frac{8}{3} & 4 \\ 4 & \frac{32}{5} \\ \end{array} \right] % \left[ \begin{array}{c} c_{1} \\ c_{2} \end{array} \right] % = % \left[ \begin{array}{r} 8 (e-2) \\ -32 (e-3) \end{array} \right], $$ which has the solution $$ \left[ \begin{array}{c} c_{1} \\ c_{2} \end{array} \right] % = % \left[ \begin{array}{c} 24 (-19 + 7 e) \\ 300 - 110 e \end{array} \right]. % $$ The following plot shows the residual error function $g(x) - c_{1}x - c_{2}x^{2}.$
