Here's a discovery i made about polynomial extrapolation (it probably can be derived from Newton's series but i haven't done it nor have i seen it done).
First, some preliminaries:
3 points can be fit by a unique quadratic polynomial, i.e., parabola. 4 points can be fit by a unique cubic polynomial.
n points can be fit by a unique (n+1)th degree polynomial.
Suppose we have the 3 points x(0)==1, x(1)==4, x(2)==9. These are obviously fit by the quadratic y(x)=x**2. Now the question is: how do you extrapolate x(3)?
Answer: Form the binomial expansion of (a-b)**3=0 and note the coefficients: Notice that I expand a CUBIC. For a cubic I would expand (a-b)**4, etc
a**3-3(a**2)b+3ab**2-b**3=0
The coefficients are 1, -3, +3, -1
The discovery I made is that these terms correspond in order to the coefficients of x(3), x(2), x(1), and x(0).
We are trying to determine x(3).
We then have: x(3) = 3x(2) -3x(1) +x(0),
or: x(3) = 3(9)-3(4)+1 = 16
This method extrapolates the next term of any polynomial where the x intervals are the same size. | [reply] |
correction to the above:
The discovery I made is that these terms correspond in order to the MULTIPLIERS needed for x(3), x(2), x(1), and x(0).
We are trying to determine x(3).
| [reply] |