Depict the value of y at x = 46 from the
following data using Newton’s Forward
interpolation formula.
x 45 50 55 60 65
y 114.84 96.16 83.32 74.48 68.48

To find the value of \( y \) at \( x = 46 \) using Newton's Forward Interpolation formula, we'll follow these steps:

### 1. Construct the Difference Table:
We begin by creating a table of forward differences for the given data.

| \(x\)  | \(y\)     | \( \Delta y \)  | \( \Delta^2 y \) | \( \Delta^3 y \) | \( \Delta^4 y \) |
|--------|-----------|-----------------|-----------------|-----------------|-----------------|
| 45     | 114.84    |                 |                 |                 |                 |
| 50     | 96.16     | -18.68          |                 |                 |                 |
| 55     | 83.32     | -12.84          | 5.84            |                 |                 |
| 60     | 74.48     | -8.84           | 4.00            | -1.84           |                 |
| 65     | 68.48     | -6.00           | 2.84            | -1.16           | 0.68            |

### 2. Apply Newton's Forward Interpolation Formula:
The formula is:

\[
y = y_0 + u \Delta y_0 + \frac{u(u-1)}

Learn how to find the value of y at x = 46 using Newton's Forward Interpolation formula. This problem involves constructing a difference table and applying Newton's method step-by-step. Suitable for high school students and anyone interested in numerical methods.

Math Problem Statement

Solution

1. Construct the Difference Table:

2. Apply Newton's Forward Interpolation Formula:

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$x$	$y$	$\Delta y$	$\Delta^2 y$	$\Delta^3 y$	$\Delta^4 y$
45	114.84
50	96.16	-18.68
55	83.32	-12.84	5.84
60	74.48	-8.84	4.00	-1.84
65	68.48	-6.00	2.84	-1.16	0.68