Q.Using Newton's backward interpolation formula on the same table ($x=0,1,2,3$; $y=2,5,10,17$), estimate $y$ at $x=2.5$.

With $x=2.5$ close to the end of the table, Newton's backward formula applies directly using the last row's own backward differences.

Setting up $p$

$x_3=3,h=1$, so $p=\dfrac{2.5-3}{1}=-0.5$.

Finding the backward differences at the last row

$\nabla y_3=y_3-y_2=17-10=7$. $\nabla^2y_3=\nabla y_3-\nabla y_2=7-5=2$ (using $\nabla y_2=y_2-y_1=10-5=5$).

Applying Newton's backward formula

$y=17+(-0.5)(7)+\frac{(-0.5)(0.5)}{2}(2)=17-3.5+\frac{-0.25}{2}(2)=17-3.5-0.25=13.25$ …

Using $x_n$ as the FIRST point instead of the LAST point of the table when computing $p=\frac{x-x_n}{h}$ for the backward formula — al …