How can a transform a polynomial to another coordinate system?

Question

Using assorted matrix math, I've solved a system of equations resulting in coefficients for a polynomial of degree 'n'

Ax^(n-1) + Bx^(n-2) + ... + Z

I then evaulate the polynomial over a given x range, essentially I'm rendering the polynomial curve. Now here's the catch. I've done this work in one coordinate system we'll call "data space". Now I need to present the same curve in another coordinate space. It is easy to transform input/output to and from the coordinate spaces, but the end user is only interested in the coefficients [A,B,....,Z] since they can reconstruct the polynomial on their own. How can I present a second set of coefficients [A',B',....,Z'] which represent the same shaped curve in a different coordinate system.

If it helps, I'm working in 2D space. Plain old x's and y's. I also feel like this may involve multiplying the coefficients by a transformation matrix? Would it some incorporate the scale/translation factor between the coordinate systems? Would it be the inverse of this matrix? I feel like I'm headed in the right direction...

Update: Coordinate systems are linearly related. Would have been useful info eh?

Answer 1

The problem statement is slightly unclear, so first I will clarify my own interpretation of it:

You have a polynomial function

f(x) = C_nxⁿ + C_n-1x^n-1 + ... + C₀

[I changed A, B, ... Z into C_n, C_n-1, ..., C₀ to more easily work with linear algebra below.]

Then you also have a transformation such as:   z = ax + b   that you want to use to find coefficients for the same polynomial, but in terms of z:

f(z) = D_nzⁿ + D_n-1z^n-1 + ... + D₀

This can be done pretty easily with some linear algebra. In particular, you can define an (n+1)×(n+1) matrix T which allows us to do the matrix multiplication

  d = T * c ,

where d is a column vector with top entry D₀, to last entry D_n, column vector c is similar for the C_i coefficients, and matrix T has (i,j)-th [i^th row, j^th column] entry t_ij given by

  t_ij = (j choose i) aⁱ b^j-i.

Where (j choose i) is the binomial coefficient, and = 0 when i > j. Also, unlike standard matrices, I'm thinking that i,j each range from 0 to n (usually you start at 1).

This is basically a nice way to write out the expansion and re-compression of the polynomial when you plug in z=ax+b by hand and use the binomial theorem.