Friday, September 18, 2009

polynomials - six point quintic

This sketch is a follow-up to my sketch about polynomials through points, which showed dynamic quadratic and cubic functions calculated from the coordinates of independent points in the plane. My approach in that sketch used matrices to solve systems of equations, but I felt I had reached the end of my tether with the inversion of a 4x4 matrix, as you will see if you "show all hidden" in the previous sketch. This sketch uses a different method that streamlines the process considerably, so I was able to plot the quintic without too much pain. Actually, the quintic took me about three minutes of tinkering with the quartic sketch, and extending to a sixth degree polynomial would be a task of similar magnitude.
I suggest that producing such a sketch would be a challenging and insightful exercise for a curious pre-calculus student. When does this polynomial fail to exist? How could we extend this construction to a higher degree? Can we employ a similar concept to find any of the roots? All of the roots?
The calculation process used in this sketch is pretty neat. I do not know what the process is called, but I do not doubt that the principle is familiar to someone. To me, this approach seems to keep the process tidier and more intuitive than would be the case with matrices, although it accomplishes a nearly equivalent task. Matrices would be handy in a single static case with numerical values, but solving a 5x5 matrix in this environment is far less practical than the method which I have employed.
We are trying to find f(x), a polynomial of degree n. Now f(x) passes through (is defined by) n+1 known points, the first of which we will call (a,f(a)). Next, we define the function fT(x)=f(x)-f(a), so fT(x) is the vertical translation of f(x) with a zero at x=a. Now (x-a) is a factor of fT(x) and our task of finding f(x) is finished if we find fT(x), one of whose linear factors we know. In this way, we recursively lower the degree of the polynomial that we seek. When finally we find ourselves seeking a linear function, the task is of little difficulty and we have only to retrace our steps and piece together our dismantled polynomial.

No comments:

Post a Comment