Question

From the data
x 1 1.5 2 2.5 3
f(x) 1 .5 1.3 .8 1.5
interpolate the value of f(2.8) using the Newton’s backward difference formula.

Answer 1

Newton’s backward difference table

x 1 1.5 2 2.5 3

f(x) 1 0.5 1.3 0.8 1.5

$∇$ -0.5 0.8 -0.5 0.7

$∇^{2}$ 1.3 -1.3 1.2

$∇^{3}$ -2.6 2.5

$∇^{4}$ 5.1

$f(x)=y_{n}​(x)+p∇y_{n}​(x)+\frac{p(p+1)∇^{2}y_{n}​(x)​}{2!}+\frac{p(p+1)(p+2)∇^{3}y_{n}​(x)}{3!​}+\frac{p(p+1)(p+2)(p+3)∇4yn​(x)}{4!}​....\\where\phantom{i}p=\frac{x−x_{0}}{h​​}\\p=\frac{2.8−3}{0.5}\\​p=−0.4,\phantom{i}y_{n}​=1.5;\phantom{i}∇y_{n}​(x)=0.7;\phantom{i}∇^{2}y_{n}​(x)=1.2;\phantom{i}∇^{3}y_{n}​(x)=2.5;∇^{4}y_{n}​(x)=5.1\\f(2.8)=1.5+(−0.4)0.7+\frac{(−0.4)(−0.4+1)1.2​}{2!}+\frac{(−0.4)(−0.4+1)(−0.4+2)2.5​}{3!}+\frac{(−0.4)(−0.4+1)(−0.4+2)(−0.4+3)5.1​}{4!}\\f(2.8)=1.5−0.28−0.144−0.16−0.21216\\f(2.8)=0.70384$