UPSC 2024 Maths Optional Paper 2 Q8b — Step-by-Step Solution

15 marks · Section B

Newton's forward difference interpolation · Numerical Analysis · asked 3× in 13 yrs · Read the full method →

Question

Using Newton’s forward difference formula, estimate $f(2.5)$ from:

$x$	1	2	3	4	5	6
$f(x)$	0	1	8	27	64	125

Build the forward-difference table; constant third differences → cubic; apply Newton’s forward formula at $p=(2.5-1)/1=1.5$ .

Step 1 — Forward-difference table.

$x$	$f$	$\Delta f$	$\Delta^2 f$	$\Delta^3 f$	$\Delta^4 f$
1	0	1	6	6	0
2	1	7	12	6	0
3	8	19	18	6
4	27	37	24
5	64	61
6	125

Third differences are constant (= 6); fourth and higher vanish. The data is the cubic $f(x)=(x-1)^3$ .

Step 2 — Newton’s forward formula.

With $x_0=1$ , $h=1$ , $p=(2.5-1)/1=1.5$ :

f(2.5)\approx f_0+p\,\Delta f_0+\frac{p(p-1)}{2}\Delta^2 f_0+\frac{p(p-1)(p-2)}{6}\Delta^3 f_0.

=0+(1.5)(1)+\frac{(1.5)(0.5)}{2}(6)+\frac{(1.5)(0.5)(-0.5)}{6}(6).

=1.5+2.25-0.375=3.375.

(This is exact since the function is cubic and the rule is exact for cubics: $(1.5)^3=3.375$ .)

\boxed{\;f(2.5)\approx 3.375.\;}