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Finite Differences and Polynomials

Date: 06/17/99 at 00:22:52
From: Chris Harris
Subject: Why are finite differences and polynomials linked?

I have read that f(x) is a polynomial in x and of degree n if and only 
if the finite differences of x go constant at the nth row of 
differences. I do not doubt the validity of this, but could you please 
explain why this should be true in general? I can't understand why it 
is so except for in the case of linear polynomials P(x) = mx+b. Or is 
this something one can't easily grasp until college?

Date: 06/17/99 at 10:07:59
From: Doctor Anthony
Subject: Re: Why are finite differences and polynomials linked?

It is easy to show why this happens.

If you have say a cubic in x such as f(x) = ax^3 

           f(1) = a
           f(2) = 8a
           f(3) = 27a
           f(4) = 64a
           f(5) = 125a

   n =    1       2      3       4        5
 f(n)=    a      8a     27a      64a     125a

1st diff.     7a    19a      37a     61a
2nd diff.        12a     18a      24a
3rd diff.             6a       6a

With 3rd differences constant.

If you wish to write out the rather tedious algebra you could make up 
a table similar to that shown above with the entries

         x          (x+1)         (x+2)          (x+3)          (x+4) 
 f(x)=  ax^3       a(x+1)^3      a(x+2)^3       a(x+3)^3       

1st diff  a(3x^2+3x+1)  a(3x^2+9x+7)  a(3x^2+15x+19)   a(3x^2+21x+37)
2nd diff            a(6x+6)      a(6x+12)         a(6x+18)        
3rd diff                     6a             6a

again 3rd differences are constant.

You can carry out a similar exercise with functions like

     f(x) = ax^3 + bx^2 + cx + d

You will find again that third differences are constant.

A theoretical proof follows along the lines below:

Define  [x]^n = x(x-1)(x-2).....(x-n+1)     (there are n factors)

so for example  [x]^3 =  x(x-1)(x-2)   and we define [x]^0 = 1

     Let D[f(x)] = f(x+1) - f(x)

and so D[[x]^n]  = [x+1]^n - [x]^n

                 = (x+1)[x]^(n-1) - [x]^(n-1)(x-n+1)

                 = [x]^(n-1) [x + 1 - x + n -1]

                 = n.[x]^(n-1) 

also    D^2[x]^n = D.D[x]^n

                 = D[n.[x]^(n-1)]

                 = n.D[x]^(n-1)

                 = n(n-1)[x]^(n-2)

which is of course the familiar result for ordinary differentiation.

Similarly  D^3[x]^n = n(n-1)(n-2)[x]^(n-3)

and if n = 3 this gives  D^3[x]^3 = n(n-1)(n-2)[x]^0

                                  = n(n-1)(n-2)

                                  = 3 x 2 x 1 =  6

So with a cubic polynomial the third differences are constant. In the 
general case an nth degree polynomial will have the nth differences 

- Doctor Anthony, The Math Forum   
Associated Topics:
High School Sequences, Series

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