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The $n^{th}$ order difference of a polynomial of degree $n$ is

- zero
- one
- some constant
- undefined

Please explain the solution.

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Best answer

**Answer is C**, i.e. **Some Constant**.

**First Order Difference is nothing but First order derivative. Similarly, ***n*^{th} order difference means n^{th }order Derivative.

Take polynomial f(x) of Order n :

(Where a_{i }is any Real Number and **a _{0 }is NOT equal to Zero**.)

First Order Difference will be : $f'(x) = a_{0}\,n\, x^{n-1} + a_1 (n-1)\, x^{n-2} + \cdot \cdot \cdot \cdot +\,\, a_{n-1}$ i.e. a polynomial of (n-1) degree.

Similarly, You can see that Second Order difference (or Derivative) will be a Polynomial of (n-2) degree. And So on, The

$n^{th}$ degree polynomial would just be a "Zero Degree" Polynomial which is :

**Which is Some Constant. **

**We can note the following :**

**The n**^{th }difference, Δ^{n}, is n! a_{0}**The n+1 difference, Δ**^{n+1}, and higher differences, are all zero.

Note That :

Zero Degree Polynomial : $f(x) = a_0\,x^{0} = a_0$ (Where $a_0$ is any Real Number and can even be Zero)

One Degree Polynomial : $f(x) = a_0\,x^{1} + a_1 = a_0 \,x + a_1$ (Where $a_i$ is any Real Number and $a_0$ Can **NOT be Zero**)

and So on.