Question

The following table lists an $n^{th}$ order polynomial f(x) =$a_n{x}^n+a_{n-1}{x}^{n-1}+...+a_{1}x+a_0$ and the forward differences evaluated at equally spaced values of x, The order of the polynomial is

x	f(x)	$\Delta$f	${\Delta }^2$f	${\Delta }^3$f
-0.4	1.7648	-0.2965	0.089	-0.03
-0.3	1.4683	-0.2075	0.059	-0.0228
-0.2	1.2608	-0.1485	0.0362	-0.0156
-0.1	1.1123	-0.1123	0.0206	-0.0084
0	1	-0.0917	0.0122	-0.0012
0.1	0.9083	-0.0795	0.011	0.006
0.2	0.8288	0.0685	0.017	0.0132

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is D 4

Clearly the entries in the fourth forward difference $(i.e,{\Delta }^{4}f)$ are all same which is 0.0072
i.e., we can say that${\Delta }^{5}f\left(x\right)=0$ (i.e., we cannot find $5^{th}$ difference)
$\because$ The order of the polynomial is 4.