Question

Using Newton's backward difference formula find the value of y at x=3.5 for following data:

x	0	1	2	3	4
y	3	2	3	6	11

Answer 1

8.25

Answer 2

The given data is:

x	0	1	2	3	4
y	3	2	3	6	11

We need to find the value of y at x = 3.5 using Newton's backward difference formula. The points are equally spaced with interval $h = 1$.

Newton's backward difference formula is given by:

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

where $x = x_n + ph$.

We choose the last point as the reference point, so $x_n = 4$ and $y_n = y_4 = 11$. The value of x at which we want to find y is 3.5. $h = 1$

Calculate $p$:

$p = \frac{x - x_n}{h} = \frac{3.5 - 4}{1} = -0.5$

Construct the backward difference table:

x	y	$\nabla y$	$\nabla^2 y$	$\nabla^3 y$	$\nabla^4 y$
0	3
1	2	2 - 3 = -1
2	3	3 - 2 = 1	1 - (-1) = 2
3	6	6 - 3 = 3	3 - 1 = 2	2 - 2 = 0
4	11	11 - 6 = 5	5 - 3 = 2	2 - 2 = 0	0 - 0 = 0

The backward differences at $x_n=x_4=4$ are the last values in each column:

$y_4 = 11$ $\nabla y_4 = 5$ $\nabla^2 y_4 = 2$ $\nabla^3 y_4 = 0$ $\nabla^4 y_4 = 0$

Since the third and fourth differences are zero, the polynomial is of degree 2, and the formula truncates after the second difference term.

Substitute the values of $y_4$, $p$, $\nabla y_4$, and $\nabla^2 y_4$ into Newton's backward difference formula:

$y(3.5) = y_4 + p \nabla y_4 + \frac{p(p+1)}{2!} \nabla^2 y_4 + \dots$ $y(3.5) = 11 + (-0.5)(5) + \frac{(-0.5)(-0.5+1)}{2} (2)$ $y(3.5) = 11 - 2.5 + \frac{(-0.5)(0.5)}{2} (2)$ $y(3.5) = 11 - 2.5 + \frac{-0.25}{2} (2)$ $y(3.5) = 11 - 2.5 - 0.25$ $y(3.5) = 8.5 - 0.25$ $y(3.5) = 8.25$

Thus, the value of y at x = 3.5 is 8.25.