The question asks to apply the Gauss forward interpolation formula to find $f(x)$ at $x = 3.5$. Given the table, here's how we approach it.

Table of Values:

Gauss Forward Interpolation Formula:

The Gauss forward formula is a Newton forward difference interpolation formula, expressed as:

$f(x) = f_0 + p \Delta f_0 + \frac{p(p-1)}{2!} \Delta^2 f_0 + \frac{p(p-1)(p-2)}{3!} \Delta^3 f_0 + \cdots$

Where:

• $p = \frac{x - x_0}{h}$
• $h$ is the uniform spacing between the $x$-values.
• $\Delta f_0, \Delta^2 f_0, \Delta^3 f_0$ are the forward differences at $x_0$.

Step 1: Calculate $p$

Given that we are to find $f(x)$ at $x = 3.5$ and the closest value in the table is $x_0 = 2.5$, the uniform step $h = 1.0$ (since the spacing between consecutive $x$-values is 1.0).

Now, we compute $p$:

$p = \frac{3.5 - 2.5}{1} = 1$

Step 2: Forward Differences

Next, we calculate the forward differences from the given data. Let me do the calculations for the forward differences and plug them into the formula to obtain the required result.Here are the calculated forward differences from the given dataHere are the calculated forward differences from the given

Step 3: Apply the Gauss Forward Formula

Using the forward differences and the formula:

$f(x) = f_0 + p \Delta f_0 + \frac{p(p-1)}{2!} \Delta^2 f_0 + \frac{p(p-1)(p-2)}{3!} \Delta^3 f_0$

Since $p = 1$, we have:

$f(3.5) = 24.364 + 1 \cdot 15.401 + \frac{1(1-1)}{2!} \cdot 26.575 + \frac{1(1-1)(1-2)}{3!} \cdot 45.143$

The terms involving higher-order differences ($\Delta^2 f_0$ and $\Delta^3 f_0$) drop out because $p(p-1) = 0$ for them.

Thus:

$f(3.5) = 24.364 + 15.401 = 39.765$

So, the value of $f(3.5)$ is approximately 39.765 based on Gauss forward interpolation.

Would you like any further explanation or calculations?

Related Questions:

1. How is the Gauss forward interpolation formula derived?
2. Can the backward formula be used to obtain the same result?
3. What would happen if the points were unevenly spaced?
4. How do higher-order terms affect accuracy in interpolation?
5. What is the difference between Newton's forward and backward formulas?

Tip: Interpolation accuracy depends on how well the given data fits the function being approximated, so including more data points may yield better results!