The following data represents the function
x
f (x)  e ,
x
1 1.5 2 2.5 3
f (x)
2.7183 4.4817 7.3891 12.1825 20.0855
Estimate the value of
f (1.75)
by using the Newton’s forward difference interpolation. Obtain a bound
on the error at
x 1.75 .

To estimate the value of \( f(1.75) \) using **Newton's forward difference interpolation**, we'll follow these steps:

### Step 1: Tabulate the given data
The data provided is:

\[
\begin{array}{c|c}
x & f(x) = e^x \\
\hline
1.0 & 2.7183 \\
1.5 & 4.4817 \\
2.0 & 7.3891 \\
2.5 & 12.1825 \\
3.0 & 20.0855 \\
\end{array}
\]

### Step 2: Forward differences
We construct the forward difference table. Denote the first forward difference by \( \Delta f(x) \), the second by \( \Delta^2 f(x) \), and so on.

\[
\begin{array}{c|c|c|c|c}
x & f(x) & \Delta f(x) & \Delta^2 f(x) & \Delta^3 f(x) \\
\hline
1.0 & 2.7183 & 1.7634 & 1.1440 & 0.0486 \\
1.5 & 4.4817 & 2.9074 & 1.1926 &  \\
2.0 & 7.3891 & 4.7934 &  \\
2.5 & 12.1825 & 7.9030 &  \\
3.0 & 20.0855 &  \\
\end{array}
\]

### Step 3: Newton's forward difference formula

Newton's forward difference formula for interpolation 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 difference between consecutive \( x \)-values (in this case \( h = 0.5 \)), and \( x_0 = 1.0 \).

For \( x = 1.75 \):
\[
p = \frac{1.75 - 1.0}{0.5} = 1.5
\]

Now, we substitute into the interpolation formula:

\[
P(1.75) = f(1.0) + p \Delta f(1.0) + \frac{p(p-1)}{2!} \Delta^2 f(1.0) + \frac{p(p-1)(p-2)}{3!} \Delta^3 f(1.0)
\]

Substituting the values:

\[
P(1.75) = 2.7183 + 1.5(1.7634) + \frac{1.5(0.5)}{2!} (1.1440) + \frac{1.5(0.5)(-0.5)}{3!} (0.0486)
\]

### Step 4: Calculation
\[
P(1.75) = 2.7183 + 2.6451 + 0.4290 - 0.0061
\]

\[
P(1.75) \approx 5.7863
\]

### Step 5: Error bound

The formula for the error bound in Newton's forward difference interpolation is:

\[
E(x) = \frac{f^{(n+1)}(\xi) (x - x_0)(x - x_1) \cdots (x - x_n)}{(n+1)!}
\]

Here, the maximum value of \( f^{(4)}(x) \) (since the 4th forward difference is small) for \( f(x) = e^x \) can be estimated as the maximum value of \( e^x \) on the interval \( [1, 3] \). Since \( e^x \) is increasing, the maximum value is at \( x = 3 \), which gives \( e^3 \approx 20.0855 \).

The error bound can be approximated as:

\[
E(1.75) \approx \frac{20.0855 (1.75 - 1.0)(1.75 - 1.5)(1.75 - 2.0)(1.75 - 2.5)}{4!}
\]

\[
E(1.75) \approx \frac{20.0855 \times 0.75 \times 0.25 \times (-0.25) \times (-0.75)}{24}
\]

\[
E(1.75) \approx \frac{20.0855 \times 0.03515625}{24} \approx 0.0294
\]

### Final result:
The estimated value of \( f(1.75) \) using Newton's forward difference interpolation is approximately **5.7863**.  
The bound on the error is approximately **0.0294**.

---

Would you like more details or clarifications on this method?

### Related Questions:
1. How is Newton's backward difference interpolation different from the forward difference?
2. What are other methods for polynomial interpolation?
3. How does the step size \( h \) affect the accuracy of Newton’s interpolation?
4. Can you derive the error term for higher-order Newton interpolations?
5. What is the difference between Lagrange interpolation and Newton's method?

**Tip:** When working with interpolation, always ensure the spacing between \( x \)-values (step size) is consistent for applying Newton’s method effectively.

The following data represents the function f(x) = e^x, x = [1, 1.5, 2, 2.5, 3], f(x) = [2.7183, 4.4817, 7.3891, 12.1825, 20.0855]. Estimate the value of f(1.75) by using the Newton’s forward difference interpolation. Obtain a bound on the error at x = 1.75.

Estimate the value of f(1.75) using Newton's forward difference interpolation for the exponential function f(x) = e^x. Detailed solution with step-by-step forward difference table and error bound calculation. Suitable for numerical methods students.

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