Solving systems of equations via Lagrange interpolation?

If you have a set of *n* points, you can find a polynomial of degree *n* or less that interpolates them using systems of equations. But Lagrange interpolation can get the job done much faster.

My question is this: Is there a way to do the first part in reverse? I.e., can systems of equations be transformed into corresponding ordered pairs in a relatively hassle-free manner, interpolated, and then the resulting coefficients become the solutions?

Example: Let's say I take the following systems of equations: a + 2b + 3c = 6, 2a - 2b + 4c = 4, and 3a - 5b + c = -1. Could I put them into a 3X4 matrix, turn it into ordered pairs that all fit the function y = ax^2 + bx + c (this is the critical part), interpolate to get that function, and then read out what a, b, and c are? Or would that take too many steps?

Re: Solving systems of equations via Lagrange interpolation?

Hey phys251.

You can indeed project one representation to another basis via Hilbert-Space theory and orthogonal polynomials.

Are you aware of these techniques?

Re: Solving systems of equations via Lagrange interpolation?

Quote:

Originally Posted by

**chiro** Hey phys251.

You can indeed project one representation to another basis via Hilbert-Space theory and orthogonal polynomials.

Are you aware of these techniques?

No, I'm not. How would that construct an alternate way to solve a system of equations?

Re: Solving systems of equations via Lagrange interpolation?

Well basically what the Hilbert-Space theory and orthogonal polynomials allow you to do is to project a signal or a function to another basis which allows you to project one thing to a quadratic and then get the co-efficients a,b,c for the best fit of that function.

You will need to determine a range of integration and derive a set of orthogonal polynomials (for a quadratic you will have three) but once you have done, then perform three integrals and you'll get your coeffecients.

If this doesn't sound like what you are asking for then I'm probably mis-understanding your question.