1. **Problem Statement:** We are given population data of a town for the years 1980, 1990, 2000, 2010, and 2020 (in thousands): 15, 30, 47, 53, and 65 respectively. We need to find the population in the year 2003 using the Gauss backward difference formula. 2. **Understanding the formula:** The Gauss backward difference formula is: $$f(u) = f(0) + u\Delta f(-1) + \frac{(u+1)u}{2!} \Delta^2 f(-1) + \frac{(u+1)u(u-1)}{3!} \Delta^3 f(-2) + \frac{(u+2)(u+1)u(u-1)}{4!} \Delta^4 f(-2) + \cdots$$ where $u = \frac{x - x_0}{h}$, $x_0$ is the base point (last known data point), $h$ is the interval between data points, and $\Delta$ denotes backward differences. 3. **Step 1: Define variables and intervals** - The data points are at years: 1980, 1990, 2000, 2010, 2020. - Interval $h = 10$ years. - We want to find population at $x = 2003$. - Choose $x_0 = 2020$ (last data point), so $$u = \frac{2003 - 2020}{10} = \frac{-17}{10} = -1.7$$ 4. **Step 2: Tabulate backward differences** Population $f$ (in thousands): \begin{align*} f_0 &= 65 \\ % 2020 f_1 &= 53 \\ % 2010 f_2 &= 47 \\ % 2000 f_3 &= 30 \\ % 1990 f_4 &= 15 % 1980 \end{align*} Calculate backward differences: \begin{align*} \Delta f_0 &= f_0 - f_1 = 65 - 53 = 12 \\ \Delta f_1 &= f_1 - f_2 = 53 - 47 = 6 \\ \Delta f_2 &= f_2 - f_3 = 47 - 30 = 17 \\ \Delta f_3 &= f_3 - f_4 = 30 - 15 = 15 \end{align*} Second backward differences: \begin{align*} \Delta^2 f_0 &= \Delta f_0 - \Delta f_1 = 12 - 6 = 6 \\ \Delta^2 f_1 &= \Delta f_1 - \Delta f_2 = 6 - 17 = -11 \\ \Delta^2 f_2 &= \Delta f_2 - \Delta f_3 = 17 - 15 = 2 \end{align*} Third backward differences: \begin{align*} \Delta^3 f_0 &= \Delta^2 f_0 - \Delta^2 f_1 = 6 - (-11) = 17 \\ \Delta^3 f_1 &= \Delta^2 f_1 - \Delta^2 f_2 = -11 - 2 = -13 \end{align*} Fourth backward difference: \begin{align*} \Delta^4 f_0 &= \Delta^3 f_0 - \Delta^3 f_1 = 17 - (-13) = 30 \end{align*} 5. **Step 3: Apply Gauss backward difference formula** Recall: $$f(u) = f_0 + u \Delta f_0 + \frac{(u+1)u}{2!} \Delta^2 f_0 + \frac{(u+1)u(u-1)}{3!} \Delta^3 f_0 + \frac{(u+2)(u+1)u(u-1)}{4!} \Delta^4 f_0$$ Substitute values: \begin{align*} f(-1.7) &= 65 + (-1.7)(12) + \frac{(-1.7+1)(-1.7)}{2} (6) + \frac{(-1.7+1)(-1.7)(-1.7-1)}{6} (17) + \frac{(-1.7+2)(-1.7+1)(-1.7)(-1.7-1)}{24} (30) \end{align*} Calculate each term: - $u \Delta f_0 = -1.7 \times 12 = -20.4$ - $\frac{(u+1)u}{2} \Delta^2 f_0 = \frac{(-0.7)(-1.7)}{2} \times 6 = \frac{1.19}{2} \times 6 = 0.595 \times 6 = 3.57$ - $\frac{(u+1)u(u-1)}{6} \Delta^3 f_0 = \frac{(-0.7)(-1.7)(-2.7)}{6} \times 17$ Calculate numerator: $$(-0.7)(-1.7) = 1.19$$ $$1.19 \times (-2.7) = -3.213$$ Divide by 6: $$\frac{-3.213}{6} = -0.5355$$ Multiply by 17: $$-0.5355 \times 17 = -9.1035$$ - $\frac{(u+2)(u+1)u(u-1)}{24} \Delta^4 f_0 = \frac{(0.3)(-0.7)(-1.7)(-2.7)}{24} \times 30$ Calculate numerator stepwise: $$0.3 \times (-0.7) = -0.21$$ $$-0.21 \times (-1.7) = 0.357$$ $$0.357 \times (-2.7) = -0.9639$$ Divide by 24: $$\frac{-0.9639}{24} = -0.0401625$$ Multiply by 30: $$-0.0401625 \times 30 = -1.204875$$ 6. **Step 4: Sum all terms** \begin{align*} f(-1.7) &= 65 - 20.4 + 3.57 - 9.1035 - 1.204875 \\ &= 65 - 20.4 + 3.57 - 9.1035 - 1.204875 \\ &= (65 - 20.4) + (3.57 - 9.1035 - 1.204875) \\ &= 44.6 + (-6.738375) \\ &= 37.861625 \end{align*} 7. **Final answer:** The estimated population in 2003 is approximately **37.86 thousand**.