Q.Using Newton's backward interpolation formula on the same table ($x=0,1,2,3,4$; $y=1,3,7,13,21$), estimate $y$ at $x=3.5$.

Since $x=3.5$ lies near the end of the table, Newton's backward formula (using $y_4$ and its own backward differences) applies directly.

Setting up $p$

Here $x_4=4$ (the last point), $h=1$, so $p=\dfrac{x-x_4}{h}=\dfrac{3.5-4}{1}=-0.5$.

Finding the backward differences at the last row

$\nabla y_4=y_4-y_3=21-13=8$. $\nabla^2y_4=\nabla y_4-\nabla y_3=8-6=2$ (using $\nabla y_3=y_3-y_2=13-7=6$ from the same table).

Applying Newton's backward formula

$y=y_4+p\nabla y_4+\frac{p(p+1)}{2!}\nabla^2y_4=21+(-0.5)(8)+\frac{(-0.5)(-0.5+1)}{2}(2)$ …

Using the FORWARD formula's $p(p-1)$ pattern instead of the backward formula's correct $p(p+1)$ pattern — the two formulas' higher-order coeffi …