1. **Problem Statement:** (a) Construct a forward difference table for given $x$ and $f(x)$ values. (b) Construct a backward difference table for $f(x) = 3x^3 - 2x^2 + x + 5$ at $x=0,1,2,3,4$. (c) Given $y$ values with $\Delta^4 y=0$, find the missing $y$ at $x=5$. 2. **Newton's Forward Interpolation:** (a) Estimate $f(0.125)$ using Newton's forward interpolation from given data. (b) Analyze sales data for TNT bars from 2017 to 2025. --- ### (a) Forward Difference Table Construction - Given $x = [0.1,0.3,0.5,0.7,0.9,1.1,1.3]$ and $f = [0.003,0.067,0.148,0.248,0.370,0.518,0.697]$. - Forward difference $\Delta f_i = f_{i+1} - f_i$. - Construct successive differences until the last column. | $x$ | $f$ | $\Delta f$ | $\Delta^2 f$ | $\Delta^3 f$ | $\Delta^4 f$ | $\Delta^5 f$ | $\Delta^6 f$ | |---|---|---|---|---|---|---|---| |0.1|0.003|0.064|0.011|0.002|0.000|0.000|0.000| |0.3|0.067|0.081|0.013|0.002|0.000|0.000| | |0.5|0.148|0.100|0.015|0.002|0.000| | | |0.7|0.248|0.122|0.017|0.002| | | | |0.9|0.370|0.148|0.019| | | | | |1.1|0.518|0.179| | | | | | |1.3|0.697| | | | | | | ### (b) Backward Difference Table for $f(x) = 3x^3 - 2x^2 + x + 5$ at $x=0,1,2,3,4$ - Calculate $f(x)$ values: $f(0)=5$, $f(1)=3(1)-2(1)+1+5=7$, $f(2)=24-8+2+5=23$, $f(3)=81-18+3+5=71$, $f(4)=192-32+4+5=169$ - Backward difference $\nabla f_i = f_i - f_{i-1}$ starting from the right. | $x$ | $f$ | $\nabla f$ | $\nabla^2 f$ | $\nabla^3 f$ | $\nabla^4 f$ | |---|---|---|---|---|---| |0|5| | | | | |1|7|2| | | | |2|23|16|14| | | |3|71|48|32|18| | |4|169|98|50|18|0| ### (c) Find missing $y$ at $x=5$ given $\Delta^4 y=0$ - Given $x = 2,3,4,5,6$ and $y = 45.0,49.2,54.1,?,67.4$ - Construct forward difference table with unknown $y_5$. - Use $\Delta^4 y=0$ to solve for missing $y$. Calculate differences: $\Delta y_2 = 49.2 - 45.0 = 4.2$ $\Delta y_3 = 54.1 - 49.2 = 4.9$ $\Delta y_4 = y_5 - 54.1$ $\Delta y_5 = 67.4 - y_5$ Second differences: $\Delta^2 y_2 = 4.9 - 4.2 = 0.7$ $\Delta^2 y_3 = (y_5 - 54.1) - 4.9 = y_5 - 59.0$ $\Delta^2 y_4 = (67.4 - y_5) - (y_5 - 54.1) = 121.5 - 2y_5$ Third differences: $\Delta^3 y_2 = (y_5 - 59.0) - 0.7 = y_5 - 59.7$ $\Delta^3 y_3 = (121.5 - 2y_5) - (y_5 - 59.0) = 180.5 - 3y_5$ Fourth difference: $\Delta^4 y_2 = (180.5 - 3y_5) - (y_5 - 59.7) = 240.2 - 4y_5$ Set $\Delta^4 y_2 = 0$: $$240.2 - 4y_5 = 0 \implies y_5 = \frac{240.2}{4} = 60.05$$ ### Question 3 (a) Newton's Forward Interpolation to estimate $f(0.125)$ - Given $x = [0,0.25,0.5,0.75,1.0]$, $f = [2,2.1,2.3,2.7,3.5]$ - Step size $h=0.25$, $x_0=0$, $p=\frac{0.125-0}{0.25}=0.5$ - Construct forward difference table: | $x$ | $f$ | $\Delta f$ | $\Delta^2 f$ | $\Delta^3 f$ | $\Delta^4 f$ | |---|---|---|---|---|---| |0|2|0.1|0.05|0.05|0.05| |0.25|2.1|0.2|0.1|0.1| | |0.5|2.3|0.4|0.2| | | |0.75|2.7|0.8| | | | |1.0|3.5| | | | | - Newton's forward interpolation formula: $$f(x) = f_0 + p\Delta f_0 + \frac{p(p-1)}{2!} \Delta^2 f_0 + \frac{p(p-1)(p-2)}{3!} \Delta^3 f_0 + \frac{p(p-1)(p-2)(p-3)}{4!} \Delta^4 f_0$$ Calculate terms: $f_0 = 2$ $p\Delta f_0 = 0.5 \times 0.1 = 0.05$ $\frac{p(p-1)}{2} \Delta^2 f_0 = \frac{0.5 \times (-0.5)}{2} \times 0.05 = -0.00625$ $\frac{p(p-1)(p-2)}{6} \Delta^3 f_0 = \frac{0.5 \times (-0.5) \times (-1.5)}{6} \times 0.05 = 0.003125$ $\frac{p(p-1)(p-2)(p-3)}{24} \Delta^4 f_0 = \frac{0.5 \times (-0.5) \times (-1.5) \times (-2.5)}{24} \times 0.05 = -0.0013021$ Sum all: $$f(0.125) = 2 + 0.05 - 0.00625 + 0.003125 - 0.0013021 = 2.0455729 \approx 2.046$$ ### Question 3 (b) Sales Data Analysis - Years: 2017, 2019, 2021, 2023, 2025 - Sales: 40, 45, 48, 50, 57 (billions) - This data can be used for trend analysis or interpolation but no specific question asked. --- **Final answers:** - (a) Forward difference table constructed as above. - (b) Backward difference table constructed with $f(x)$ values and differences. - (c) Missing $y$ at $x=5$ is $\boxed{60.05}$. - (3a) Estimated $f(0.125) = \boxed{2.046}$.