1. **Problem Statement:** (a) Estimate $f(0.125)$ using Newton's forward interpolation formula given the data: $$\begin{array}{c|ccccc} x & 0 & 0.25 & 0.5 & 0.75 & 1.0 \\ \hline f(x) & 2 & 2.1 & 2.3 & 2.7 & 3.5 \end{array}$$ (b) Estimate sales for years 2020 and 2022 using Newton's forward interpolation with the data: $$\begin{array}{c|ccccc} \text{Year} & 2017 & 2019 & 2021 & 2023 & 2025 \\ \hline \text{Sales} & 40 & 43 & 48 & 52 & 57 \end{array}$$ 2. **Formula:** Newton's forward interpolation formula is: $$P_n(x) = f(x_0) + p\Delta f(x_0) + \frac{p(p-1)}{2!}\Delta^2 f(x_0) + \frac{p(p-1)(p-2)}{3!}\Delta^3 f(x_0) + \cdots$$ where $p = \frac{x - x_0}{h}$ and $h$ is the equal spacing between $x$ values. 3. **Part (a) Calculation:** - Given $x_0=0$, $h=0.25$, and $x=0.125$, calculate $p = \frac{0.125-0}{0.25} = 0.5$. - Construct forward difference table for $f(x)$: $$\begin{array}{c|ccccc} x & 0 & 0.25 & 0.5 & 0.75 & 1.0 \\ f(x) & 2 & 2.1 & 2.3 & 2.7 & 3.5 \\\Delta f & 0.1 & 0.2 & 0.4 & 0.8 \\\Delta^2 f & 0.1 & 0.2 & 0.4 \\\Delta^3 f & 0.1 & 0.2 \\\Delta^4 f & 0.1 \end{array}$$ - Apply formula: $$P_4(0.125) = 2 + 0.5(0.1) + \frac{0.5(0.5-1)}{2}(0.1) + \frac{0.5(0.5-1)(0.5-2)}{6}(0.1) + \frac{0.5(0.5-1)(0.5-2)(0.5-3)}{24}(0.1)$$ - Calculate each term: - $0.5 \times 0.1 = 0.05$ - $\frac{0.5 \times (-0.5)}{2} \times 0.1 = -0.0125$ - $\frac{0.5 \times (-0.5) \times (-1.5)}{6} \times 0.1 = 0.00625$ - $\frac{0.5 \times (-0.5) \times (-1.5) \times (-2.5)}{24} \times 0.1 = -0.00390625$ - Sum all terms: $$2 + 0.05 - 0.0125 + 0.00625 - 0.00390625 = 2.03984375$$ 4. **Part (b) Calculation:** - Years: 2017, 2019, 2021, 2023, 2025 with $h=2$ years. - For 2020, $p = \frac{2020 - 2017}{2} = 1.5$; for 2022, $p = \frac{2022 - 2017}{2} = 2.5$. - Forward difference table for sales: $$\begin{array}{c|ccccc} \text{Year} & 2017 & 2019 & 2021 & 2023 & 2025 \\ \text{Sales} & 40 & 43 & 48 & 52 & 57 \\\Delta & 3 & 5 & 4 & 5 \\\Delta^2 & 2 & -1 & 1 \\\Delta^3 & -3 & 2 \\\Delta^4 & 5 \end{array}$$ - For 2020 ($p=1.5$): $$P_4(2020) = 40 + 1.5(3) + \frac{1.5(0.5)}{2}(2) + \frac{1.5(0.5)(-0.5)}{6}(-3) + \frac{1.5(0.5)(-0.5)(-1.5)}{24}(5)$$ - Calculate terms: - $1.5 \times 3 = 4.5$ - $\frac{1.5 \times 0.5}{2} \times 2 = 0.75$ - $\frac{1.5 \times 0.5 \times (-0.5)}{6} \times (-3) = 0.1875$ - $\frac{1.5 \times 0.5 \times (-0.5) \times (-1.5)}{24} \times 5 = 0.1171875$ - Sum: $$40 + 4.5 + 0.75 + 0.1875 + 0.1171875 = 45.5546875$$ - For 2022 ($p=2.5$): $$P_4(2022) = 40 + 2.5(3) + \frac{2.5(1.5)}{2}(2) + \frac{2.5(1.5)(0.5)}{6}(-3) + \frac{2.5(1.5)(0.5)(-0.5)}{24}(5)$$ - Calculate terms: - $2.5 \times 3 = 7.5$ - $\frac{2.5 \times 1.5}{2} \times 2 = 3.75$ - $\frac{2.5 \times 1.5 \times 0.5}{6} \times (-3) = -0.9375$ - $\frac{2.5 \times 1.5 \times 0.5 \times (-0.5)}{24} \times 5 = -0.1953125$ - Sum: $$40 + 7.5 + 3.75 - 0.9375 - 0.1953125 = 49.1171875$$ **Final answers:** - (a) $f(0.125) \approx 2.0398$ - (b) Sales in 2020 $\approx 45.55$ billion, Sales in 2022 $\approx 49.12$ billion