$k$th order backward difference reduces $k$th degree polynomial

Question

$X_t$ is a degree $k$ polynomial at $t$. How to deduce that $\triangledown ^kX_t$ is a constant?

Answer 1

Let us write $$ X_t = a_0 + a_1 t + a_2 t^2 + \dots + a_{k-1} t^{k-1} + a_k t^k, $$ where $a_j, \, j = 0, \dots, k$ are the polynomial coefficients of $X_t$. Now, $\nabla$ is a linear operator, i.e. $$ \nabla X_t = \nabla a_0 + \nabla a_1 t + \nabla a_2 t^2 + \dots + \nabla a_{k-1} t^{k-1} + \nabla a_k t^k. $$ Note that for the $j$th term in this expression we have $$ \nabla a_j t^j = a_j t^j - a_j (t-1)^j = a_j t^j - a_j \left[ t^j - \binom{j}{1} t^{j-1} + \binom{j}{2} t^{j-2} - \dots \right] = j a_j t^{j-1} + \dots $$ where we have used the binomial theorem to expand $(t-1)^j$. Therefore, $\nabla$ has the effect of reducing the order of each term in $X_t$ by one. Consequently, $\nabla X_t$ is a polynomial of order $k-1$. Similarly, $$ \nabla^2 X_t = \nabla (\nabla X_t) \text{ is a polynomial of degree } k-2, $$ $$\nabla^3 X_t = \nabla(\nabla^2 X_t) \text{ is a polynomial of degree } k-3, $$ $$ \vdots $$ $$ \nabla^k X_t = \nabla(\nabla^{k-1} X_t) \text{ is a polynomial of degree } k - k = 0, $$ i.e. it is a constant.