Q.Using Newton's forward interpolation formula on $x=0,1,2,3$; $y=2,5,10,17$, estimate $y$ at $x=0.5$.

With $x=0.5$ close to the start of the table, Newton's forward formula applies directly using $y_0$ and its own differences.

$p=0.5$; $y=2+0.5(3)+\frac{0.5(-0.5)}2(2)=3.25$.

Setting up $p$

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

Applying Newton's forward formula (using $y_0=2,\Delta y_0=3,\Delta^2y_0=2$ from the previous exercise's table)

Check (independent recomputation via the known underlying function $y=x^2+2x+2$): $f(0.5)=(0.5)^2+2(0.5)+2=0.25+1+2=3.25$ — matches exactly.

Forgetting to divide by $2!=2$ in the second-difference term — the factorial denominator grows with each successive term of the series and is easy to drop when working quickly.