1. **State the problem:** We are given values of a function $f(x)$ at points $x = 0.12, 0.16, 0.20, 0.24, 0.28, 0.32$ and corresponding $f(x)$ values. We need to find $f(0.14)$ using Newton-Gregory finite difference interpolation. 2. **Formula used:** For equally spaced data points, the Newton-Gregory forward interpolation formula is: $$ P(x) = f(x_0) + p\Delta f(x_0) + \frac{p(p-1)}{2!}\Delta^2 f(x_0) + \frac{p(p-1)(p-2)}{3!}\Delta^3 f(x_0) + \cdots $$ where $p = \frac{x - x_0}{h}$, $h$ is the spacing between $x$ values, and $\Delta$ denotes forward differences. 3. **Calculate $h$ and $p$:** $$ h = 0.16 - 0.12 = 0.04 $$ $$ p = \frac{0.14 - 0.12}{0.04} = \frac{0.02}{0.04} = 0.5 $$ 4. **Construct the forward difference table:** | $x$ | $f(x)$ | $\Delta f$ | $\Delta^2 f$ | $\Delta^3 f$ | $\Delta^4 f$ | $\Delta^5 f$ | |------|---------|------------|--------------|--------------|--------------|--------------| |0.12 | 0.6144 | 0.0112 | 0.0016 | 0.0004 | 0 | 0 | |0.16 | 0.6256 | 0.0144 | 0.0020 | 0 | 0 | | |0.20 | 0.6400 | 0.0168 | 0.0020 | 0 | | | |0.24 | 0.6576 | 0.0192 | 0.0020 | | | | |0.28 | 0.6784 | 0.0224 | | | | | |0.32 | 0.7024 | | | | | | Calculations: - $\Delta f(0.12) = 0.6256 - 0.6144 = 0.0112$ - $\Delta f(0.16) = 0.6400 - 0.6256 = 0.0144$ - $\Delta f(0.20) = 0.6576 - 0.6400 = 0.0168$ - $\Delta f(0.24) = 0.6784 - 0.6576 = 0.0192$ - $\Delta f(0.28) = 0.7024 - 0.6784 = 0.0224$ - $\Delta^2 f(0.12) = 0.0144 - 0.0112 = 0.0032$ - $\Delta^2 f(0.16) = 0.0168 - 0.0144 = 0.0024$ - $\Delta^2 f(0.20) = 0.0192 - 0.0168 = 0.0024$ - $\Delta^2 f(0.24) = 0.0224 - 0.0192 = 0.0032$ - $\Delta^3 f(0.12) = 0.0024 - 0.0032 = -0.0008$ - $\Delta^3 f(0.16) = 0.0024 - 0.0024 = 0$ - $\Delta^3 f(0.20) = 0.0032 - 0.0024 = 0.0008$ - $\Delta^4 f(0.12) = 0 - (-0.0008) = 0.0008$ - $\Delta^4 f(0.16) = 0.0008 - 0 = 0.0008$ - $\Delta^5 f(0.12) = 0.0008 - 0.0008 = 0$ 5. **Apply the interpolation formula:** $$ P(0.14) = f(0.12) + p\Delta f(0.12) + \frac{p(p-1)}{2!} \Delta^2 f(0.12) + \frac{p(p-1)(p-2)}{3!} \Delta^3 f(0.12) + \frac{p(p-1)(p-2)(p-3)}{4!} \Delta^4 f(0.12) + \frac{p(p-1)(p-2)(p-3)(p-4)}{5!} \Delta^5 f(0.12) $$ Calculate each term: - $f(0.12) = 0.6144$ - $p\Delta f(0.12) = 0.5 \times 0.0112 = 0.0056$ - $\frac{p(p-1)}{2} \Delta^2 f(0.12) = \frac{0.5 \times (0.5 - 1)}{2} \times 0.0032 = \frac{0.5 \times (-0.5)}{2} \times 0.0032 = -0.0004$ - $\frac{p(p-1)(p-2)}{6} \Delta^3 f(0.12) = \frac{0.5 \times (-0.5) \times (-1.5)}{6} \times (-0.0008) = \frac{0.375}{6} \times (-0.0008) = -0.00005$ - $\frac{p(p-1)(p-2)(p-3)}{24} \Delta^4 f(0.12) = \frac{0.5 \times (-0.5) \times (-1.5) \times (-2.5)}{24} \times 0.0008 = \frac{-0.9375}{24} \times 0.0008 = -0.00003125$ - $\frac{p(p-1)(p-2)(p-3)(p-4)}{120} \Delta^5 f(0.12) = 0$ (since $\Delta^5 f(0.12) = 0$) 6. **Sum all terms:** $$ P(0.14) = 0.6144 + 0.0056 - 0.0004 - 0.00005 - 0.00003125 + 0 = 0.61951875 $$ 7. **Final answer:** $$ f(0.14) \approx 0.6195 $$