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example of under-determined polynomial interpolation∗ rmilson† 2013-03-21 13:59:24 Consider the following interpolation problem: Given x1 , y1 , x2 , y2 ∈ R with x1 6= x2 to determine all cubic polynomials p(x) = ax3 + bx2 + cx + d, x, a, b, c, d ∈ R such that p(x1 ) = y1 , p(x2 ) = y2 . This is a linear problem. Let P3 denote the vector space of cubic polynomials. The underlying linear mapping is the multi-evaluation mapping E : P3 → R 2 , given by p 7→ p(x1 ) , p(x2 ) p ∈ P3 The interpolation problem in question is represented by the equation y1 E(p) = y2 where p ∈ P3 is the unknown. One can recast the problem into the traditional form by taking standard bases of P3 and R2 and then seeking all possible a, b, c, d ∈ R such that a 3 2 (x1 ) (x1 ) x1 1 b y1 = 3 2 y2 (x2 ) (x2 ) x2 1 c d ∗ hExampleOfUnderdeterminedPolynomialInterpolationi created: h2013-03-21i by: hrmilsoni version: h32840i Privacy setting: h1i hExamplei h15A06i † This text is available under the Creative Commons Attribution/Share-Alike License 3.0. You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. 1 However, it is best to treat this problem at an abstract level, rather than mucking about with row reduction. The Lagrange interpolation formula gives us a particular solution, namely the linear polynomial p0 (x) = x − x1 x − x2 y1 + y2 , x2 − x1 x1 − x2 x∈R The general solution of our interpolation problem is therefore given as p0 + q, where q ∈ P3 is a solution of the homogeneous problem E(q) = 0. A basis of solutions for the latter is, evidently, q1 (x) = (x − x1 )(x − x2 ), q2 (x) = xq1 (x), x∈R The general solution to our interpolation problem is therefore p(x) = x − x2 x − x1 y1 + y2 + (ax + b)(x − x1 )(x − x2 ), x2 − x1 x1 − x2 x ∈ R, with a, b ∈ R arbitrary. The general under-determined interpolation problem is treated in an entirely analogous manner. 2