To solve the problem using Newton's Backward Interpolation Formula, I'll first transcribe the data from the image and then proceed with the calculation.

Transcription of Data

The table seems to have the following structure:

x	y (f(x))	Δy	Δ²y	Δ³y	Δ⁴y
65	68.49
60	74.49	-6.00
55	83.33	-8.84	2.84
50	96.16	-12.83	4.00	1.16
45	114.84	-18.68	5.84	1.84	0.68

Now let's use the Newton's Backward Interpolation formula:

Newton's Backward Interpolation Formula

Given the data points $(x_n, y_n)$, Newton's Backward Interpolation formula is:

$P(x) = y_n + u \Delta y_n + \frac{u(u+1)}{2!} \Delta^2 y_n + \frac{u(u+1)(u+2)}{3!} \Delta^3 y_n + \cdots$

Where:

• $u = \frac{x - x_n}{h}$
• $h$ is the difference between the $x$ values, which is constant.

Step 1: Determine $u$

If we are interpolating for a value $x = 46$:

$u = \frac{46 - 45}{5} = 0.2$

For $x = 63$:

$u = \frac{63 - 65}{5} = -0.4$

Step 2: Apply the Newton's Backward Formula

For $x = 46$:

$P(46) = 114.84 + 0.2(-18.68) + \frac{0.2(0.2 + 1)}{2} \times 5.84 + \frac{0.2(0.2 + 1)(0.2 + 2)}{6} \times 1.84 + \cdots$

For $x = 63$:

$P(63) = 68.49 + (-0.4)(-6) + \frac{(-0.4)(-0.4 + 1)}{2} \times 2.84 + \frac{(-0.4)(-0.4 + 1)(-0.4 + 2)}{6} \times 1.16 + \cdots$

Step 3: Calculate the Values

Let's substitute and calculate:

For $x = 46$:

$P(46) = 114.84 + (-3.736) + (0.5)(0.1168) + \cdots$

$P(46) \approx 114.84 - 3.736 + 0.584 + 0.0245 \approx 111.7125$

For $x = 63$:

$P(63) = 68.49 + 2.4 + 0.568 + 0.31 \approx 71.768$

Thus, the interpolated values are approximately:

• $P(46) \approx 111.7125$
• $P(63) \approx 71.768$

Would you like further details on this or have any questions?

Related Questions:

1. What is the difference between Newton's Forward and Backward Interpolation formulas?
2. How can the error in interpolation be estimated?
3. What is the significance of choosing an appropriate interpolation method?
4. Could you apply the same method for different values of $x$?
5. How does the value of $u$ affect the interpolation result?

Tip: Newton's Backward Interpolation is more accurate near the end of the data range, while Forward Interpolation is more accurate near the beginning.