Q.Using Newton's forward interpolation formula on the table $x=0,1,2,3,4$; $y=1,3,7,13,21$, estimate $y$ at $x=0.5$.

Since $x=0.5$ lies near the start of the table, Newton's forward formula (using $y_0$ and its own differences) applies directly.

$p=0.5$; $y=1+0.5(2)+\frac{0.5(-0.5)}2(2)=1.75$.

Setting up $p$

Here $x_0=0$, $h=1$, so $p=\dfrac{x-x_0}{h}=\dfrac{0.5-0}{1}=0.5$.

Applying Newton's forward formula (using $y_0=1,\Delta y_0=2,\Delta^2y_0=2$ from the previous worked example's table; $\Delta^3y_0=0$ ends the series)

Check (independent recomputation via the known underlying function $y=x^2+x+1$, confirmed in the previous example): $f(0.5)=(0.5)^2+0.5+1=0.25+0.5+1=1.75$ — matches the interpolated value exactly.

Miscomputing $p(p-1)$ for a fractional $p$ (e.g. computing $0.5(0.5-1)$ as a positive number instead of the correct $0.5\times(-0.5)=-0.25$) — carry the negative sign through carefully when $p<1$.