Polynomial Functions and Vanishing Triangles

	The Calculus of Finite Differences, sometimes called the method
of vanishing triangles, is a simple procedure for determining which
polynomial function f(n) will fit a sequence of given numerical data of
the form "f(0), f(1), f(2), ...".  The approach to constructing the
triangle is just the reverse of that for Pascal's Triangle, that is,
instead of adding subsequent entries in a row, you now subtract
consecutive terms.

	An example will make my point clear.  Let f(n) = 2n2 + 3n + 4.
The sequence of values for n = 0, 1, 2, 3, ... is 4, 9, 18, 31, ... .
The vanishing triangle is

			4     9     18     31
			   5     9     13
			      4     4
			         0

(Any zeros obtained are superfluous and can be omitted in practice.
They are only mentioned to illustrate how the term "triangle" is
involved.)

	If one does several functions like the example (i.e. 2nd-
degree functions), it will be noticed that:

	1) the value in the 3rd row is always a constant and
	   twice the coefficient of n2,

	2) the first entry in the 2nd row equals the sum of the
	   coefficients of n2 and n, and

	3) the first entry in the first row is the constant term
	   in f(n).

Hence, if at least the first three values of f(n), beginning with
f(0), are known, the coefficients can be determined by examining the
first entries in the three rows.



	That the foregoing is true is easily proved.  Let f(n) =
an2 + bn + c.  Therefore, the triangle formed with the first four
values is

		c     a+b+c     4a+2b+c     9a+3b+c
		  a+b       3a+b           5a+b
		        2a            2a

The expressions "2a, a+b, and c" confirm the earlier statements as true.



	This will give the student who likes to discover formulas for
himself a tool to do so.  For example, what is the general formula
for the triangular numbers?  The sequence is 1, 3, 6, 10, ... .
Though this function begins with the value for n = 1, once the triangle
is set up, it's a simple matter to "back it up" one value.

			0     1     3     6     10
			   1     2     3     4
			      1     1     1

As 2a = 1, a + b = 1, and c = 0, the coefficients are a = b = ½  and
c = 0. Therefore, f(n) = ½ n2 + n + 0 = ½ n(n + 1)

	The utility of this process will be seen as one continues
working with numerical functions, especially those of higher degree.
For example, for data requiring a 3rd-degree function (i.e. f(n) =
an3 + bn2 + cn + d), the necessary expressions are "6a, 6a + 2b,
a + b + c, and d."  Determining the "guide" expressions for higher
degree functions just requires careful work in the triangle.

	This is only a brief examination into this topic, as I prefer
to leave any further discoveries up to you, the reader.




This originally appeared in the Bulletin of the Kansas Association
of Teachers of Matematics.  February 1971, p.20.