Equation of a Sequence with Constant Third Differences

Date: 05/26/98 at 12:20:43
From: Sohail Malik
Subject: the method of difference

Dear Dr. Math, 
 
I am fully capable of using the method of difference when the sequence 
has a second difference. But I have a sequence that has a third 
difference:

   1   4   10   20   35   56 .... 

Please could you tell me the method of difference when there is a 
third difference? I would be eternally grateful if you could help me 
with this problem. 

Yours faithfully,

Sohail Malik

Date: 05/26/98 at 16:03:51
From: Doctor Anthony
Subject: Re: the method of difference

Given your sequence, you can find out the following information:

             n =   1     2      3       4        5       6  ...
          f(n) =   1     4     10      20       35      56  ...
1st difference D      3     6      10       15      21      ...
2nd difference D^2       3      4       5        6          ...
3rd difference D^3           1      1       1               ...

With constant 3rd differences, the expression for f(n) will be a cubic 
polynomial in n:

   f(n) =  an^3 + bn^2 + cn + d

Substituting in for n, you find:

   f(1) =   a +   b +  c + d =  1             
   f(2) =  8a +  4b + 2c + d =  4
   f(3) = 27a +  9b + 3c + d = 10
   f(4) = 64a + 16b + 4c + d = 20

And you have 4 equations to solve for the 4 unknowns a, b, c, and d.

There is another way, depending on the Gregory-Newton formula, which 
in this case would be:

   f(n) = f(1) + C(n-1,1)D(1) + C(n-1,2)D^2(1) + C(n-1,3)D^3(1)

      = 1 + (n - 1)(3) + (n - 1)(n - 2)/2 (3) 
          + (n - 1)(n - 2)(n - 3)/6 (1)

      = 1 + 3n - 3 + (3/2)(n^2 - 3n + 2) + (1/6)(n^3 - 6n^2 + 11n - 6)

      = (1/6)n^3 + (1/2)n^2 + (1/3)n 

We can check this with n = 1 and n = 2:

   f(1) = 1  and  f(2) = 4  

Thus the equation is giving the correct values.  

-Doctor Anthony, The Math Forum
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