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Date: 01/21/2002 at 14:31:51
From: Monica Kerai

Here's what I understand so far:

a+b+c: 5  8  12  17  23 ...
3a+b:   3   4   5   6 ...
2b:     1   1   1 ...

You figure out what a, b, and c are, and then you put them into this
equation:  an^2 + bn + c.

Where do I get the (a+b+c) (3a +b) (2b) part from?
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Date: 01/22/2002 at 14:08:41
From: Doctor Peterson

Hi, Monica.

I think you meant 2a, not 2b. As you'll see, you're more likely to
remember it right if you see where it comes from.

What you are doing is taking successive differences in a sequence.
When they all turn out to be zero, you know that the sequence is a
polynomial of that degree; in this case, the third difference will be
zero, so the sequence represents a quadratic equation.

Suppose you have an arbitrary quadratic sequence

x[n] = an^2 + bn + c

Then your differences will look like this:

n=1        n=2          n=3

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

3a+b        5a+b

2a

Those (on the left) are the expressions you were told to use.

You can simplify the work a bit by separating out each power, since if
you add two sequences together, the differences will be the sum of the
differences of the two sequences. Look at these cases:

an^2      +       bn        +        c
a  4a  9a 16a     b  2b  3b  4b     c   c   c   c
3a  5a  7a         b   b   b         0   0   0
2a  2a             0   0
0

So the first difference row will start with 3a+b, and the second
difference row with 2a. The constant term (c) has no effect on any
differences; the linear term (b) drops out after the first difference,
and so on. That helps in remembering or reconstructing the formula.

See what you get if you add on a cubic term.

We have some discussions of the method of finite differences in our
archives, which may help clarify the process; start here:

Method of Finite Differences
http://mathforum.org/dr.math/problems/gillett.10.12.00.html

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Polynomials
High School Sequences, Series

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