Question

Using Newton's backward difference formula, find y at x = 3.5 from following data:

Answer 1

13.240625

Answer 2

The given data points are:

x	0	1	2	3	4	5
y	5.2	8	10.4	12.4	14	15.2

We need to find y at x = 3.5 using Newton's backward difference formula. The points are equally spaced with interval $h = 1$. Since we are interpolating near the end of the table (x=3.5 is closer to x=5 than x=0), Newton's backward difference formula is suitable. The formula is:

$y(x) = y_n + p \nabla y_n + \frac{p(p+1)}{2!} \nabla^2 y_n + \frac{p(p+1)(p+2)}{3!} \nabla^3 y_n + \dots$

where $x = x_n + ph$.

We take the last point as the reference point, so $x_n = 5$ and $y_n = y_5 = 15.2$. We need to find y at $x = 3.5$.

$p = \frac{x - x_n}{h} = \frac{3.5 - 5}{1} = -1.5$

Now, we construct the backward difference table:

x	y	$\nabla y$	$\nabla^2 y$	$\nabla^3 y$	$\nabla^4 y$	$\nabla^5 y$
0	5.2
1	8.0	2.8
2	10.4	2.4	-0.4
3	12.4	2.0	-0.4	0
4	14.0	1.6	-0.4	0.4
5	15.2	1.2	-0.4	0	-0.4

The backward differences at $x_n=x_5=5$ are the last values in each column:

$y_5 = 15.2$ $\nabla y_5 = 1.2$ $\nabla^2 y_5 = -0.4$ $\nabla^3 y_5 = 0$ $\nabla^4 y_5 = -0.4$ $\nabla^5 y_5 = 0$

Substitute these values and $p = -1.5$ into Newton's backward difference formula:

$y(3.5) = y_5 + p \nabla y_5 + \frac{p(p+1)}{2!} \nabla^2 y_5 + \frac{p(p+1)(p+2)}{3!} \nabla^3 y_5 + \frac{p(p+1)(p+2)(p+3)}{4!} \nabla^4 y_5 + \dots$

$y(3.5) = 15.2 + (-1.5)(1.2) + \frac{(-1.5)(-1.5+1)}{2} (-0.4) + \frac{(-1.5)(-1.5+1)(-1.5+2)}{6} (0) + \frac{(-1.5)(-1.5+1)(-1.5+2)(-1.5+3)}{24} (-0.4) + \dots$

$y(3.5) = 15.2 + (-1.5)(1.2) + \frac{(-1.5)(-0.5)}{2} (-0.4) + 0 + \frac{(-1.5)(-0.5)(0.5)(1.5)}{24} (-0.4) + \dots$

$y(3.5) = 15.2 - 1.8 + \frac{0.75}{2} (-0.4) + \frac{0.5625}{24} (-0.4) + \dots$

$y(3.5) = 15.2 - 1.8 + 0.375 (-0.4) + 0.0234375 (-0.4) + \dots$

$y(3.5) = 15.2 - 1.8 - 0.15 - 0.009375 + \dots$

$y(3.5) = 13.4 - 0.15 - 0.009375$

$y(3.5) = 13.25 - 0.009375$

$y(3.5) = 13.240625$