1. We are given two sets of data points and two models for approximation by the least-squares method: - Model 1: $$\varphi(x) = c_1 \frac{1}{x} + c_2$$ with points $$(-3, 32.3), (-2.5, -45.4), (-1.5, 9.7), (-1, 44.9), (0.5, -21.2)$$ - Model 2: $$\varphi(x) = c_1 e^x + c_2 \frac{1}{x}$$ with points $$(-6.5, -23.6), (-6, 49.9), (-5.5, -28.9), (-5, -0.2), (-3.5, -21)$$ 2. For each model, we want to find coefficients $$c_1$$ and $$c_2$$ that minimize the sum of squared residuals: $$S = \sum_i \left(y_i - \varphi(x_i)\right)^2$$ 3. For Model 1, define basis functions: $$f_1(x) = \frac{1}{x}, \quad f_2(x) = 1$$ The system to solve is: $$\begin{bmatrix} \sum f_1(x_i)^2 & \sum f_1(x_i) f_2(x_i) \\ \sum f_1(x_i) f_2(x_i) & \sum f_2(x_i)^2 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} \sum y_i f_1(x_i) \\ \sum y_i f_2(x_i) \end{bmatrix}$$ 4. Calculate sums for Model 1: $$\sum f_1(x_i)^2 = \sum \left(\frac{1}{x_i}\right)^2 = \frac{1}{9} + \frac{1}{6.25} + \frac{1}{2.25} + 1 + 4 = 0.1111 + 0.16 + 0.4444 + 1 + 4 = 5.7155$$ $$\sum f_1(x_i) f_2(x_i) = \sum \frac{1}{x_i} = -\frac{1}{3} - \frac{1}{2.5} - \frac{1}{1.5} - 1 + 2 = -0.3333 - 0.4 - 0.6667 - 1 + 2 = -0.4$$ $$\sum f_2(x_i)^2 = 5$$ (since $$f_2(x) = 1$$ for all 5 points) $$\sum y_i f_1(x_i) = 32.3 \times (-\frac{1}{3}) + (-45.4) \times (-\frac{1}{2.5}) + 9.7 \times (-\frac{1}{1.5}) + 44.9 \times (-1) + (-21.2) \times 2 = -10.7667 + 18.16 - 6.4667 - 44.9 - 42.4 = -86.373$$ $$\sum y_i f_2(x_i) = 32.3 - 45.4 + 9.7 + 44.9 - 21.2 = 20.3$$ 5. Solve the linear system: $$\begin{bmatrix} 5.7155 & -0.4 \\ -0.4 & 5 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} -86.373 \\ 20.3 \end{bmatrix}$$ Using Cramer's rule or matrix inversion: Determinant $$D = 5.7155 \times 5 - (-0.4) \times (-0.4) = 28.5775 - 0.16 = 28.4175$$ $$c_1 = \frac{\begin{vmatrix} -86.373 & -0.4 \\ 20.3 & 5 \end{vmatrix}}{D} = \frac{-86.373 \times 5 - (-0.4) \times 20.3}{28.4175} = \frac{-431.865 + 8.12}{28.4175} = \frac{-423.745}{28.4175} \approx -14.91$$ $$c_2 = \frac{\begin{vmatrix} 5.7155 & -86.373 \\ -0.4 & 20.3 \end{vmatrix}}{D} = \frac{5.7155 \times 20.3 - (-0.4) \times (-86.373)}{28.4175} = \frac{116.0 - 34.55}{28.4175} = \frac{81.45}{28.4175} \approx 2.87$$ 6. So Model 1 approximation is: $$\boxed{\varphi(x) = -14.91 \frac{1}{x} + 2.87}$$ 7. For Model 2, basis functions are: $$f_1(x) = e^x, \quad f_2(x) = \frac{1}{x}$$ We calculate sums: $$\sum f_1(x_i)^2 = \sum e^{2x_i} = e^{-13} + e^{-12} + e^{-11} + e^{-10} + e^{-7} \approx 2.26 \times 10^{-6} + 6.14 \times 10^{-6} + 1.67 \times 10^{-5} + 4.54 \times 10^{-5} + 9.12 \times 10^{-4} = 0.00097$$ $$\sum f_1(x_i) f_2(x_i) = \sum e^{x_i} \frac{1}{x_i} = e^{-6.5}(-\frac{1}{6.5}) + e^{-6}(-\frac{1}{6}) + e^{-5.5}(-\frac{1}{5.5}) + e^{-5}(-\frac{1}{5}) + e^{-3.5}(-\frac{1}{3.5}) \approx -0.00011 - 0.00012 - 0.00018 - 0.00027 - 0.0029 = -0.00358$$ $$\sum f_2(x_i)^2 = \sum \left(\frac{1}{x_i}\right)^2 = \frac{1}{42.25} + \frac{1}{36} + \frac{1}{30.25} + \frac{1}{25} + \frac{1}{12.25} = 0.0237 + 0.0278 + 0.0331 + 0.04 + 0.0816 = 0.2062$$ $$\sum y_i f_1(x_i) = \sum y_i e^{x_i} = -23.6 e^{-6.5} + 49.9 e^{-6} - 28.9 e^{-5.5} - 0.2 e^{-5} - 21 e^{-3.5} \approx -0.0026 + 0.091 + -0.008 + -0.0013 - 0.61 = -0.53$$ $$\sum y_i f_2(x_i) = \sum y_i \frac{1}{x_i} = -23.6(-\frac{1}{6.5}) + 49.9(-\frac{1}{6}) + -28.9(-\frac{1}{5.5}) + -0.2(-\frac{1}{5}) + -21(-\frac{1}{3.5}) = 3.63 - 8.32 + 5.26 + 0.04 + 6 = 6.61$$ 8. Solve the system: $$\begin{bmatrix} 0.00097 & -0.00358 \\ -0.00358 & 0.2062 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} -0.53 \\ 6.61 \end{bmatrix}$$ Determinant: $$D = 0.00097 \times 0.2062 - (-0.00358)^2 = 0.0002 - 0.000013 = 0.000187$$ $$c_1 = \frac{\begin{vmatrix} -0.53 & -0.00358 \\ 6.61 & 0.2062 \end{vmatrix}}{D} = \frac{-0.53 \times 0.2062 - (-0.00358) \times 6.61}{0.000187} = \frac{-0.109 + 0.024}{0.000187} = \frac{-0.085}{0.000187} \approx -454.01$$ $$c_2 = \frac{\begin{vmatrix} 0.00097 & -0.53 \\ -0.00358 & 6.61 \end{vmatrix}}{D} = \frac{0.00097 \times 6.61 - (-0.00358) \times (-0.53)}{0.000187} = \frac{0.0064 - 0.0019}{0.000187} = \frac{0.0045}{0.000187} \approx 24.06$$ 9. Model 2 approximation is: $$\boxed{\varphi(x) = -454.01 e^x + 24.06 \frac{1}{x}}$$ Summary: - Model 1: $$\varphi(x) = -14.91 \frac{1}{x} + 2.87$$ - Model 2: $$\varphi(x) = -454.01 e^x + 24.06 \frac{1}{x}$$