Doubt in forward difference operator in numerical Methods

Question

To find $\triangle^3(1-2x)(1-3x)(1-4x)$

In my textbook it is given as - $P_n(x) = a_0+a_1x+a_2x^2+....+a_nx^n$ be n degree polynomial.

THen $ \triangle^kP_n(x) = 0, k>n; \triangle^nP_n(x) = a_0(n!)$

But how to prove this?

Answer 1

Use another polynomial basis, $P_n(x)=b_0+b_1x+b_2\binom{x}2+b_3\binom{x}3+...+b_n\binom{x}{n}$. As only the last term contains the $n$th power of $x$, one finds $b_n=n!a_n$. Further \begin{align} △\binom{x}k&=\binom{x+1}k-\binom{x}k\\ &=\frac{(x+1)x(x-1)...(x-k+2)-x(x-1)(x-2)...(x-k+1)}{k!}\\ &=\binom{x}{k-1}\frac{(x+1)-(x-k+1)}{k}=\binom{x}{k-1} \end{align} so that $$ △P_n(x)=b_1+b_2\binom{x}1+b_3\binom{x}2+...+b_n\binom{x}{n-1},\\ △^mP_n(x)=b_m+b_{m+1}\binom{x}1+b_{m+2}\binom{x}2+...+b_n\binom{x}{n-m}.\\ $$