To solve the problem:

(1) Construct Newton's Forward Difference Table

We begin by constructing the difference table for the given data:

Given Data:

Step 1: First Differences ($\Delta y$):

$\Delta y_i = y_{i+1} - y_i$ $\Delta y_0 = 6 - 3 = 3, \quad \Delta y_1 = 11 - 6 = 5, \quad \Delta y_2 = 18 - 11 = 7, \quad \Delta y_3 = 27 - 18 = 9$

Step 2: Second Differences ($\Delta^2 y$):

$\Delta^2 y_i = \Delta y_{i+1} - \Delta y_i$ $\Delta^2 y_0 = 5 - 3 = 2, \quad \Delta^2 y_1 = 7 - 5 = 2, \quad \Delta^2 y_2 = 9 - 7 = 2$

Step 3: Third Differences ($\Delta^3 y$):

$\Delta^3 y_i = \Delta^2 y_{i+1} - \Delta^2 y_i$ $\Delta^3 y_0 = 2 - 2 = 0, \quad \Delta^3 y_1 = 2 - 2 = 0$

Final Forward Difference Table:

(2) Newton's Forward Interpolation Formula

Newton's forward interpolation formula is: $f(x) = f(x_0) + u\Delta f(x_0) + \frac{u(u-1)}{2!} \Delta^2 f(x_0) + \frac{u(u-1)(u-2)}{3!} \Delta^3 f(x_0)$ where: $u = \frac{x - x_0}{h}, \quad h = x_{i+1} - x_i$

Step 1: Given values:

• $x_0 = 0$, $f(x_0) = 3$
• $h = 1$
• $x = 2.5$ $u = \frac{2.5 - 0}{1} = 2.5$

Step 2: Substitute values:

$f(2.5) = 3 + 2.5(3) + \frac{2.5(2.5-1)}{2!}(2) + \frac{2.5(2.5-1)(2.5-2)}{3!}(0)$

Simplify each term:

• $f(x_0) = 3$
• $u\Delta f(x_0) = 2.5 \cdot 3 = 7.5$
• $\frac{u(u-1)}{2!} \Delta^2 f(x_0) = \frac{2.5(1.5)}{2}(2) = \frac{3.75}{2}(2) = 3.75$
• $\frac{u(u-1)(u-2)}{3!} \Delta^3 f(x_0) = 0$

Step 3: Add the terms:

$f(2.5) = 3 + 7.5 + 3.75 = 14.25$

Final Answer:

• Forward Difference Table: As shown above.
• Approximation of $f(2.5)$: $f(2.5) = 14.25$

Would you like me to verify further, or explain specific parts of the solution?
Here are 5 related questions to expand your understanding:

1. How is Newton's forward difference formula different from backward interpolation?
2. Why do higher differences ($\Delta^3 y$) vanish for polynomial functions of degree 3?
3. What would happen if the interval $h$ is not constant?
4. How would we solve this problem using Lagrange interpolation?
5. What is the significance of the forward difference table in numerical analysis?

Tip: For accurate interpolation, ensure the value of $x$ lies within the range of given data points!