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

  1. zero
  2. one
  3. some constant
  4. undefined

Please explain the solution.

in Set Theory & Algebra
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Answer is C, i.e. Some Constant

First Order Difference is nothing but First order derivative. Similarly, nth order difference means nth  order Derivative. 

Take polynomial f(x) of Order n : 

(Where ais any Real Number and  ais 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 :

  1. The nth difference, Δn,   is n! a0
  2. The n+1 difference, Δn+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.

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