1. **Problem statement:** Estimate $f(2.15)$ and $f(2.9)$ using Newton's finite difference interpolation based on the given data points: $$\begin{array}{c|c} x & f(x) \\\hline 2.0 & 0.30103 \\ 2.2 & 0.30202 \\ 2.4 & 0.38021 \\ 2.6 & 0.41497 \\ 2.8 & 0.44716 \\ 3.0 & 0.47717 \end{array}$$ 2. **Formula and rules:** Newton's forward difference interpolation formula is: $$P(x) = f(x_0) + s\Delta f(x_0) + \frac{s(s-1)}{2!}\Delta^2 f(x_0) + \frac{s(s-1)(s-2)}{3!}\Delta^3 f(x_0) + \cdots$$ where $s = \frac{x - x_0}{h}$, $h$ is the uniform spacing between $x$ values, and $\Delta$ denotes forward differences. 3. **Calculate $h$ and $s$ values:** $h = 2.2 - 2.0 = 0.2$ For $x=2.15$, $s = \frac{2.15 - 2.0}{0.2} = 0.75$ For $x=2.9$, $s = \frac{2.9 - 2.0}{0.2} = 4.5$ 4. **Compute forward differences:** First differences $\Delta f(x)$: $\Delta f(2.0) = 0.30202 - 0.30103 = 0.00099$ $\Delta f(2.2) = 0.38021 - 0.30202 = 0.07819$ $\Delta f(2.4) = 0.41497 - 0.38021 = 0.03476$ $\Delta f(2.6) = 0.44716 - 0.41497 = 0.03219$ $\Delta f(2.8) = 0.47717 - 0.44716 = 0.03001$ Second differences $\Delta^2 f(x)$: $\Delta^2 f(2.0) = 0.07819 - 0.00099 = 0.07720$ $\Delta^2 f(2.2) = 0.03476 - 0.07819 = -0.04343$ $\Delta^2 f(2.4) = 0.03219 - 0.03476 = -0.00257$ $\Delta^2 f(2.6) = 0.03001 - 0.03219 = -0.00218$ Third differences $\Delta^3 f(x)$: $\Delta^3 f(2.0) = -0.04343 - 0.07720 = -0.12063$ $\Delta^3 f(2.2) = -0.00257 - (-0.04343) = 0.04086$ $\Delta^3 f(2.4) = -0.00218 - (-0.00257) = 0.00039$ Fourth differences $\Delta^4 f(x)$: $\Delta^4 f(2.0) = 0.04086 - (-0.12063) = 0.16149$ $\Delta^4 f(2.2) = 0.00039 - 0.04086 = -0.04047$ Fifth differences $\Delta^5 f(x)$: $\Delta^5 f(2.0) = -0.04047 - 0.16149 = -0.20196$ 5. **Estimate $f(2.15)$ using Newton's forward formula:** $$P(2.15) = f(2.0) + s\Delta f(2.0) + \frac{s(s-1)}{2!} \Delta^2 f(2.0) + \frac{s(s-1)(s-2)}{3!} \Delta^3 f(2.0) + \frac{s(s-1)(s-2)(s-3)}{4!} \Delta^4 f(2.0)$$ Calculate each term: $s = 0.75$ $s(s-1) = 0.75 \times (0.75 - 1) = 0.75 \times (-0.25) = -0.1875$ $s(s-1)(s-2) = -0.1875 \times (0.75 - 2) = -0.1875 \times (-1.25) = 0.234375$ $s(s-1)(s-2)(s-3) = 0.234375 \times (0.75 - 3) = 0.234375 \times (-2.25) = -0.52734375$ Now plug in values: $$P(2.15) = 0.30103 + 0.75 \times 0.00099 + \frac{-0.1875}{2} \times 0.07720 + \frac{0.234375}{6} \times (-0.12063) + \frac{-0.52734375}{24} \times 0.16149$$ Calculate each term: $0.75 \times 0.00099 = 0.0007425$ $\frac{-0.1875}{2} \times 0.07720 = -0.09375 \times 0.07720 = -0.007234$ $\frac{0.234375}{6} \times (-0.12063) = 0.0390625 \times (-0.12063) = -0.004711$ $\frac{-0.52734375}{24} \times 0.16149 = -0.0219727 \times 0.16149 = -0.00355$ Sum all terms: $0.30103 + 0.0007425 - 0.007234 - 0.004711 - 0.00355 = 0.28628$ (approx) 6. **Estimate $f(2.9)$ using Newton's backward difference formula:** Since $2.9$ is closer to the end, use backward differences with $x_5=3.0$: Backward differences $\nabla f(x)$: $\nabla f(3.0) = f(3.0) - f(2.8) = 0.47717 - 0.44716 = 0.03001$ $\nabla f(2.8) = 0.44716 - 0.41497 = 0.03219$ $\nabla f(2.6) = 0.41497 - 0.38021 = 0.03476$ $\nabla f(2.4) = 0.38021 - 0.30202 = 0.07819$ $\nabla f(2.2) = 0.30202 - 0.30103 = 0.00099$ Second backward differences $\nabla^2 f(x)$: $\nabla^2 f(3.0) = 0.03001 - 0.03219 = -0.00218$ $\nabla^2 f(2.8) = 0.03219 - 0.03476 = -0.00257$ $\nabla^2 f(2.6) = 0.03476 - 0.07819 = -0.04343$ $\nabla^2 f(2.4) = 0.07819 - 0.00099 = 0.07720$ Third backward differences $\nabla^3 f(x)$: $\nabla^3 f(3.0) = -0.00218 - (-0.00257) = 0.00039$ $\nabla^3 f(2.8) = -0.00257 - (-0.04343) = 0.04086$ $\nabla^3 f(2.6) = -0.04343 - 0.07720 = -0.12063$ Fourth backward differences $\nabla^4 f(x)$: $\nabla^4 f(3.0) = 0.00039 - 0.04086 = -0.04047$ $\nabla^4 f(2.8) = 0.04086 - (-0.12063) = 0.16149$ Fifth backward differences $\nabla^5 f(x)$: $\nabla^5 f(3.0) = -0.04047 - 0.16149 = -0.20196$ 7. **Newton's backward interpolation formula:** $$P(x) = f(x_n) + s \nabla f(x_n) + \frac{s(s+1)}{2!} \nabla^2 f(x_n) + \frac{s(s+1)(s+2)}{3!} \nabla^3 f(x_n) + \frac{s(s+1)(s+2)(s+3)}{4!} \nabla^4 f(x_n)$$ where $s = \frac{x - x_n}{h}$ and $x_n = 3.0$ For $x=2.9$, $s = \frac{2.9 - 3.0}{0.2} = -0.5$ Calculate terms: $s = -0.5$ $s(s+1) = -0.5 \times 0.5 = -0.25$ $s(s+1)(s+2) = -0.25 \times 1.5 = -0.375$ $s(s+1)(s+2)(s+3) = -0.375 \times 2.5 = -0.9375$ Plug in values: $$P(2.9) = 0.47717 + (-0.5) \times 0.03001 + \frac{-0.25}{2} \times (-0.00218) + \frac{-0.375}{6} \times 0.00039 + \frac{-0.9375}{24} \times (-0.04047)$$ Calculate each term: $-0.5 \times 0.03001 = -0.015005$ $\frac{-0.25}{2} \times (-0.00218) = -0.125 \times (-0.00218) = 0.0002725$ $\frac{-0.375}{6} \times 0.00039 = -0.0625 \times 0.00039 = -0.0000244$ $\frac{-0.9375}{24} \times (-0.04047) = -0.03906 \times (-0.04047) = 0.00158$ Sum all terms: $0.47717 - 0.015005 + 0.0002725 - 0.0000244 + 0.00158 = 0.46399$ (approx) **Final answers:** $$f(2.15) \approx 0.28628$$ $$f(2.9) \approx 0.46399$$