Question

Nth order divided difference of a polynomial of degree n is

Answer 1

Answer:zero

Step-by-step explanation:

Answer 2

Answer:

If $f(x)$ is a polynomial of degree $n$ then the $n^{\text {cn }}$ divided difference of $f(x)$ is a

Step-by-step explanation:

If $f(x)$ is a polynomial of degree $n$ then the $n^{\text {cn }}$ divided difference of $f(x)$ is a

constant.

Proof:

The first divided difference of $f(x)=x^{n}$

$\begin{aligned}&f\left(x_{0}, x_{1}\right)=\frac{\left(x_{0}+h\right)^{n}-x_{0}^{n}}{x_{0}+h-x_{0}} \\&f\left(x_{0}, x_{1}\right)=\frac{x_{0}^{n}+n h x_{0}^{n-1}+\ldots . . h^{n}-x_{0}^{n}}{h} \\&f\left(x_{0}, x_{1}\right)=\frac{n h x_{0}^{n-1}+\ldots . h^{n}}{h}\end{aligned}$

The second divided difference can be given as:

$f\left(x_{0}, x_{1}, x_{2}\right)=\frac{f\left(x_{1}, x_{2}\right)-f\left(x_{0}, x_{1}\right)}{x_{2}-x_{0}}$

The $\mathrm{k}^{\text {th }}$ divided difference can be given as:

$f\left(x_{0}, x_{1}, x_{2}, \ldots x_{k}\right)=\frac{f\left(x_{1}, x_{2}, \ldots x_{k}\right)-f\left(x_{0}, x_{1}, \ldots x_{k-1}\right)}{x_{k}-x_{0}}$

The divided difference operator is a linear operator.

The first divided difference is an $(n-1)^{\text {th }}$ degree polynomial.

$\therefore$ The second divided difference is an $(n-2)^{\text {th }}$ degree polynomial and hence the $n^{\text {th }}$ divided difference is $(n-n)^{\text {th }}$ degree polynomial that is a constant.