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Asked: May 10, 20262026-05-10T17:08:12+00:00 2026-05-10T17:08:12+00:00

Using assorted matrix math, I’ve solved a system of equations resulting in coefficients for

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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?

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  1. 2026-05-10T17:08:12+00:00Added an answer on May 10, 2026 at 5:08 pm

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

    You have a polynomial function

    f(x) = Cnxn + Cn-1xn-1 + … + C0

    [I changed A, B, … Z into Cn, Cn-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) = Dnzn + Dn-1zn-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 Dn, column vector c is similar for the Ci coefficients, and matrix T has (i,j)-th [ith row, jth column] entry tij given by

      tij = (j choose i) ai bj-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.

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