I was reading a difference equation textbook and came across a problem. The question asks to compute ${\Delta}^nt^{\underline3}$ for $n=1,2,3,...$, where $t^{\underline3}$ is the falling factorial "$t$ to the $3$ falling" and ${\Delta}^n$ is the $n$-th order difference operator.
However, there is a theorem that says ${\Delta}^nt^{\underline{r}}=rt^\underline{r-1}$. Therefore If I am right, had if been $n$ is known (say 2), the solution would have been as easy as
${\Delta}^2t^{\underline3}={\Delta}3t^{\underline2}=6t^{\underline1}=6t$
How will the solution be for $n=1,2,3,...$ please?I really wish someone will help.
since $t^{\underline3}=t(t-1)(t-2)$, it's aisy to see that ${\Delta}^nt^{\underline3}=0$ if $n \geq 4$.
for $n=3$, we have: ${\Delta}^3t^{\underline3}=6$